\(ĐKXĐ:x\ge2\)

\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x^2+2x-3}+\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{\left(x-1\right)\left(x+3\right)}\)

\(+\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}\)

\(-\sqrt{x-2}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)

\(TH1:\sqrt{x-2}-\sqrt{x+3}=0\Leftrightarrow\sqrt{x-2}=\sqrt{x+3}\)

\(\Leftrightarrow x-2=x+3\left(L\right)\)

\(TH2:\sqrt{x-1}-1=0\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)(t/m đk)

Vậy x = 2

\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\)

\(\Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)

Dễ thấy \(VT>0\Rightarrow3x-5>0\Leftrightarrow x>\frac{5}{3}\)

\(pt\Leftrightarrow\left(\sqrt{x^2+5}-3\right)-\left(\sqrt{x^2+12}-4\right)+3x-6=0\)

\(\Leftrightarrow\frac{x^2-4}{\sqrt{x^2+5}+3}-\frac{x^2-4}{\sqrt{x^2+12}+4}+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{\sqrt{x^2+5}+3}-\frac{x+2}{\sqrt{x^2+12}+4}+3\right)=0\)

Ta có: \(\frac{x+2}{\sqrt{x^2+5}+3}-\frac{x+2}{\sqrt{x^2+12}+4}\)\(=\left(x+2\right)\left(\frac{1}{\sqrt{x^2+5}+3}-\frac{1}{\sqrt{x^2+12}+4}\right)\)

\(=\left(x+2\right).\frac{\sqrt{x^2+12}-\sqrt{x^2+5}+1}{\left(\sqrt{x^2+5}+3\right)\left(\sqrt{x^2+12}+4\right)}>0\forall x>\frac{5}{3}\)

\(\Rightarrow x-2=0\Leftrightarrow x=2\)

Vậy x = 2

chiệu

\(ĐKXĐ:x\ge-1\)

\(14\sqrt{x+35}+6\sqrt{x+1}=84+\sqrt{x^2+36x+35}\)

Đặt \(\sqrt{x+35}=u,\sqrt{x+1}=v\)

Lúc đó \(uv=\sqrt{x^2+36x+35}\)

\(pt\Leftrightarrow14u+6v=84+uv\)

\(\Leftrightarrow\left(uv-14u\right)+\left(84-6v\right)=0\)

\(\Leftrightarrow u\left(v-14\right)-6\left(v-14\right)=0\)

\(\Leftrightarrow\left(u-6\right)\left(v-14\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}u-6=0\\v-14=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}u=6\\v=14\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+35}=\sqrt{36}\\\sqrt{x+1}=\sqrt{196}\end{cases}}\)(Vì  ẩn phụ \(u=\sqrt{x+35},v=\sqrt{x+1}\)(theo cách đặt))

\(\Leftrightarrow\orbr{\begin{cases}x+35=36\\x+1=196\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=195\end{cases}}\)(tm đk)

Vậy pt có 2 nghiệm 1 và 195

\(ĐKXĐ:x\ge-6\)

\(\sqrt{x+9}+2012\sqrt{x+6}=2012+\sqrt{\left(x+9\right)\left(x+6\right)}\)

Đặt \(\sqrt{x+9}=m\); \(\sqrt{x+6}=n\)

\(pt\Leftrightarrow m+2012n=2012+mn\)

\(\Leftrightarrow\left(mn-m\right)+\left(2012-2012n\right)=0\)

\(\Leftrightarrow m\left(n-1\right)-2012\left(n-1\right)=0\)

\(\Leftrightarrow\left(m-2012\right)\left(n-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}m-2012=0\\n-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=2012\\n=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+9}=\sqrt{4048144}\\\sqrt{x+6}=\sqrt{1}\end{cases}}\)(Vì ẩn phụ \(m=\sqrt{x+9}\);\(n=\sqrt{x+6}\)(theo cách đặt))

\(\Leftrightarrow\orbr{\begin{cases}x+9=4048144\\x+6=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4048135\\x=-5\end{cases}}\)(t/m đk)

Vậy pt có 2 nghiệm 4048135 và -5

Bài 20:\(x^4+\sqrt{x^2+2019}=2019\)(ĐKXĐ:\(\forall x\in R\))

Đặt \(\sqrt{x^2+2019}=a^2\Rightarrow a^4=x^2+2019\Rightarrow a^4-x^2=2019\)

Ta có hệ phương trình \(\hept{\begin{cases}x^4+a^2=2019\\a^4-x^2=2019\end{cases}}\)

\(\Leftrightarrow x^4+a^2=a^4-x^2\)

\(\Leftrightarrow\left(x^2-a^2\right)\left(x^2+a^2\right)+a^2+x^2=0\)

\(\Leftrightarrow\left(x^2+a^2\right)\left(x^2-a^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+a^2=0\\x^2-a^2=1\end{cases}}\)

TH1:\(x^2+a^2=0\)

Do \(VT\ge0\)\(\Rightarrow\hept{\begin{cases}a^2=0\\x^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\varnothing\\x=0\end{cases}}\left(L\right)\)

TH2:\(x^2-a^2=1\)

\(\Leftrightarrow a^4-2019-a^2=1\)(\(a^4-2019=x^2\))

\(\Leftrightarrow a^4-a^2-2020=0\)

Đặt \(a^2=t\)(\(t\ge0\))

\(\Rightarrow t^2-t-2020=0\)

\(\Delta=1-4.\left(-2020\right)=8081\)

\(\Rightarrow\orbr{\begin{cases}t=\frac{1+\sqrt{8081}}{2}\\t=\frac{1-\sqrt{8081}}{2}\left(t\ge0\right)\end{cases}}\)

Vậy \(a^2=\frac{1+\sqrt{8081}}{2}\Rightarrow\sqrt{x^2+2019}=\frac{1+\sqrt{8081}}{2}\)

Tự giải x nha

Bài 4:\(x^2+2x+7=\left(x+3\right)\sqrt{x^2+5}\)(ĐKXĐ:mọi x thuộc R)

ĐK để phương trình có nghiệm :x>-3

Đặt \(\sqrt{x^2+5}=t\)

pt\(\Leftrightarrow t^2+2x+2-\left(x+3\right)t=0\)

\(\Leftrightarrow t^2-\left(x+3\right)t+2x+2=0\)

\(\Delta=\left(x+3\right)^2-4\left(2x+2\right)=x^2+6x+9-8x-8=x^2-2x+1=\left(x-1\right)^2\)

\(\Rightarrow\orbr{\begin{cases}t=\frac{x+3-\sqrt{\left(x-1\right)^2}}{2}=2\\t=\frac{x+3+\sqrt{\left(x-1\right)^2}}{2}=\frac{2x+2}{2}=x+1\end{cases}}\)

TH1:t=2\(\Leftrightarrow\sqrt{x^2+5}=2\Leftrightarrow x^2=-1\left(L\right)\)

TH2"\(t=x+1\)\(\Leftrightarrow\sqrt{x^2+5}=\left(x+1\right)\Leftrightarrow x^2+2x+1=x^2+5\Leftrightarrow2x-4=0\Leftrightarrow x=2\left(L\right)\)

Vậy S=rỗng

Câu 20 

\(x+4+\sqrt{x^2+2019}=2019\)

\(x^4+\sqrt{x^2}=0\)

\(x^4=\sqrt{x^2}\)

\(\sqrt{x+8}=x^2\)

\(\Rightarrow x=0,1or-1\)

\(ĐKXĐ:x\inℝ\)

\(x^4+\sqrt{x^2+2019}=2019\)(1)

Đặt \(\sqrt{x^2+2019}=a^2\Rightarrow a^4=x^2+2019\Rightarrow\hept{\begin{cases}a^4-x^2=2019\\a^4-2019=x^2\end{cases}}\)

và \(\left(1\right)\Leftrightarrow x^4+a^2=2019\)

Suy ra được\(x^4+a^2=a^4-x^2\)( = 2019)

\(\Leftrightarrow x^4-a^4+a^2+x^2=0\)

\(\Leftrightarrow\left(x^2+a^2\right)\left(x^2-a^2\right)+a^2+x^2=0\)

\(\Leftrightarrow\left(x^2+a^2\right)\left(x^2-a^2+1\right)=0\)

TH1: \(x^2+a^2=0\)(2) 

Mà \(\hept{\begin{cases}a^2\ge0\forall a\inℝ\\x^2\ge0\forall x\inℝ\end{cases}}\)nên \(x^2+a^2\ge0\forall a,x\inℝ\)

\(\Rightarrow\)(2) xảy ra\(\Leftrightarrow\hept{\begin{cases}a^2=0\\x^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\x=0\end{cases}}\)(Loại vì lúc đó \(\sqrt{x^2+2019}=\sqrt{2019}=a^2=0\)(vô lí))

TH2: \(x^2-a^2+1=0\)

\(\Leftrightarrow a^4-2019-a^2+1=0\)

\(\Leftrightarrow a^4-a^2-2018=0\)(pt trùng phương)(3)

Đặt \(a^2=t\left(ĐK:t\ge0\right)\)

\(\left(3\right)\Leftrightarrow t^2-t-2018=0\)

Ta có: \(\Delta=1^2+4.2018=8073,\sqrt{\Delta}=\sqrt{8073}=3\sqrt{897}\)

\(\Rightarrow\orbr{\begin{cases}t=\frac{1+3\sqrt{897}}{2}\\t=\frac{1-3\sqrt{897}}{2}\end{cases}}\)

Dễ thấy \(1< 3\sqrt{897}\Rightarrow\frac{1-3\sqrt{897}}{2}< 0\)

Mà \(t\ge0\)nên \(t=\frac{1+3\sqrt{897}}{2}\)

Lúc đó: \(a^2=\frac{1+3\sqrt{897}}{2}\)(ẩn phụ t = a² )

hay \(\sqrt{x^2+2019}=\)\(\frac{1+3\sqrt{897}}{2}\)

Bình phương hai vế: \(x^2+2019=\frac{8074+6\sqrt{897}}{4}\)

\(\Rightarrow x^2=\frac{3\sqrt{897}-1}{2}\)

\(\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{3\sqrt{897}-1}{2}}\\x=-\sqrt{\frac{3\sqrt{897}-1}{2}}\end{cases}}\)

Vậy pt có 2 nghiệm \(\sqrt{\frac{3\sqrt{897}-1}{2}};-\sqrt{\frac{3\sqrt{897}-1}{2}}\)

\(ĐKXĐ:x\inℝ\)

\(x^2+2x+7=\left(x+3\right)\sqrt{x^2+5}\)

Đặt \(\sqrt{x^2+5}=t\)

\(pt\Leftrightarrow t^2+2x+2=\left(x+3\right)t\)

\(\Leftrightarrow t^2-\left(x+3\right)t+\left(2x+2\right)=0\)

Ta có: \(\Delta=\left(x+3\right)^2-4\left(2x+2\right)=x^2+6x+9-8x-8=\left(x-1\right)^2\)

\(\Rightarrow\orbr{\begin{cases}t=\frac{x+3-\sqrt{\left(x-1\right)^2}}{2}\\t=\frac{x+3+\sqrt{\left(x-1\right)^2}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=2\\t=x+1\end{cases}}\)

TH1: t = 2\(\Rightarrow\sqrt{x^2+5}=2\Leftrightarrow x^2+5=4\)(bình phương 2 vế)

\(\Leftrightarrow x^2=-1\left(L\right)\)

TH2: \(t=x+1\)

\(\Rightarrow\sqrt{x^2+5}=x+1\Leftrightarrow x^2+5=x^2+2x+1\)(bình phương 2 vế)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)

Vậy x = 2

\(ĐKXĐ:x\inℝ\)

\(x^2+3x+8=\left(x+5\right)\sqrt{x^2+x+2}\)

Đặt \(\sqrt{x^2+x+2}=u\)

\(pt\Leftrightarrow u^2+2x+6=\left(x+5\right)u\)

\(\Leftrightarrow u^2-\left(x+5\right)u+\left(2x+6\right)\)

Ta có: \(\Delta=\left(x+5\right)^2-4\left(2x+6\right)=x^2+10x+5-8x-24=\left(x+1\right)^2\)

\(\Rightarrow\orbr{\begin{cases}u=\frac{x+5+\sqrt{\left(x+1\right)^2}}{2}\\u=\frac{x+5-\sqrt{\left(x+1\right)^2}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}u=x+3\\u=2\end{cases}}\)

TH1: u = 2\(\Rightarrow\sqrt{x^2+x+2}=2\)\(\Leftrightarrow x^2+x+2=4\)(bình phương hai vế)

\(\Leftrightarrow x^2+x-2=0\)

Ta có: \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\\frac{-1-3}{2}=-2\end{cases}}\)(tm)

TH2: u = x + 3\(\Rightarrow\sqrt{x^2+x+2}=x+3\)

\(\Leftrightarrow x^2+x+2=x^2+6x+9\)(bình phương hai vế)

\(\Leftrightarrow5x+7=0\Leftrightarrow x=\frac{-7}{5}\)(tm)

Vậy tập nghiệm \(S=\left\{1;-2;\frac{-7}{5}\right\}\)

Còn câu nào chưa được giải nữa không ta?

\(ĐKXĐ:x\ge0\)

\(\sqrt{2x+1}+3\sqrt{4x^2-2x+1}=3+\sqrt{8x^3+1}\)

Đặt \(\sqrt{2x+1}=u\); \(\sqrt{4x^2-2x+1}=v\)

Lúc đó \(uv=\sqrt{8x^3+1}\)

\(pt\Leftrightarrow u+3v=3+uv\)

\(\Leftrightarrow\left(uv-u\right)+\left(3-3v\right)=0\)

\(\Leftrightarrow u\left(v-1\right)-3\left(v-1\right)=0\)

\(\Leftrightarrow\left(u-3\right)\left(v-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}u-3=0\\v-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}u=3\\v=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x+1}=\sqrt{9}\\\sqrt{4x^2-2x+1}=\sqrt{1}\end{cases}}\)(Theo cách đặt ẩn phụ \(u=\sqrt{2x+1}\);\(v=\sqrt{4x^2-2x+1}\))

\(\Leftrightarrow\orbr{\begin{cases}2x+1=9\\4x^2-2x+1=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=8\\4x^2-2x=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\2x\left(2x-1\right)=0\end{cases}}\)

Vậy x = 4 hoặc x= 0 hoặc \(x=\frac{1}{2}\)

\(ĐKXĐ:x\ge\sqrt{3}\)

\(2x^2+4x-8=\left(2x+3\right)\sqrt{x^2-3}\)

\(\Leftrightarrow x^2+2x-4=\left(x+\frac{3}{2}\right)\sqrt{x^2-3}\)(chia hai vế cho 2)

Đặt \(\sqrt{x^2-3}=t\)

\(pt\Leftrightarrow t^2+2x-1=\left(x+\frac{3}{2}\right)t\)

\(\Leftrightarrow t^2-\left(x+\frac{3}{2}\right)t+\left(2x-1\right)=0\)

Ta có: \(\Delta=\left(x+\frac{3}{2}\right)^2-4\left(2x-1\right)=x^2+3x+\frac{9}{4}-8x+4\)

\(=\left(x-\frac{5}{2}\right)^2\)

\(\orbr{\begin{cases}t=\frac{x+\frac{3}{2}+\sqrt{\left(x-\frac{5}{2}\right)^2}}{2}\\t=\frac{x+\frac{3}{2}-\sqrt{\left(x-\frac{5}{2}\right)^2}}{2}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}t=x-\frac{1}{2}\\t=2\end{cases}}\)

TH1: \(t=x-\frac{1}{2}\)\(\Rightarrow\sqrt{x^2-3}=x-\frac{1}{2}\)

\(\Rightarrow x^2-3=x^2-x+\frac{1}{4}\)(bình phương hai vế )

\(\Leftrightarrow x=\frac{1}{4}+3=\frac{13}{4}\)(t/m)

TH2: t = 2\(\Rightarrow\sqrt{x^2-3}=2\Rightarrow x^2-3=4\)(bình phương hai vế )

\(\Rightarrow x^2=7\Leftrightarrow x=\pm\sqrt{7}\)

Vậy pt có 3 nghiệm là \(\frac{13}{4};\pm\sqrt{7}\)

Bổ sung ĐKXĐ của bài \(\hept{\begin{cases}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{cases}}\)

\(\sqrt{2x^2-2x}=x^4-2x^3+x\)

\(\Leftrightarrow\sqrt{2x\left(x-1\right)}=x\left(x^3-2x^2+1\right)\)(\(ĐKXĐ:\orbr{\begin{cases}x\le0\\x\ge1\end{cases}}\))

\(\Leftrightarrow\sqrt{2x\left(x-1\right)}=x\left[\left(x-1\right)\left(x^2+x+1\right)-2\left(x-1\right)\right]\)

\(\Leftrightarrow\sqrt{2x\left(x-1\right)}=x\left(x-1\right)\left(x^2+x-1\right)\)

\(\Leftrightarrow\sqrt{x\left(x-1\right)}\left[\left(x^2+x-1\right)\sqrt{x\left(x-1\right)}-\sqrt{2}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x\left(x-1\right)}=0\left(1\right)\\\left(x^2+x-1\right)\sqrt{x\left(x-1\right)}=\sqrt{2}\left(2\right)\end{cases}}\)

Từ\(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x=0\left(n\right)\\x=1\left(n\right)\end{cases}}\)

Từ \(\left(2\right)\Leftrightarrow\left(x^2+x-1\right)^2x\left(x-1\right)=2\)

\(\Leftrightarrow\left(x^2+x-1\right)^2\left(x^2-x\right)=2\)

\(\Leftrightarrow\left[\left(x+1\right)^2-\left(x+1\right)-1\right]^2\left[\left(x+1\right)^2-\left(x+1\right)-2x\right]=2\)

Đặt: \(a=\left(x+1\right)^2-\left(x+1\right)\)

\(\left(2\right)\Leftrightarrow\left(a-2x\right)\left(a-1\right)^2=2\)

\(\Leftrightarrow\left(a-2x\right)\left(a^2-2a+1\right)\)

\(\Leftrightarrow\left(a^2-2x\right)\left(a^2-2x-1\right)=2\)

Đặt tiếp \(t=a^2-2x=\left(x+1\right)^2-\left(x+1\right)-2x=x^2-x\)

\(\left(2\right)\Leftrightarrow t\left(t-1\right)=2\)

\(\Leftrightarrow t^2-t-2=0\)

Có:\(\Delta=\left(-1\right)^2-4\cdot\left(-2\right)=9>0\)pt có 2 nghiệm pb

\(\sqrt{\Delta}=3\)

\(\orbr{\begin{cases}t_1=\frac{1+3}{2}=2\\t_2=\frac{1-3}{2}=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-x-2=0\\x^2-x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\left(n\right)\\x=2\left(n\right)\end{cases}}\)

Vậy \(S=\left\{-1;0;1;2\right\}\)

P/S: k kiểm lại nếu có sai xót mn check nha!

Bài 3 ĐKXĐ tự tìm

\(PT\Leftrightarrow\sqrt{x\left(6-x\right)}-2x\left(6-x\right)+15=0\)

Đặt: \(\sqrt{x\left(6-x\right)}=a\)(\(a\ge0\))

\(PT\Leftrightarrow-2a^2+a+15=0\)

Xài CT nghiệm pt bậc 2 giải a rồi tiếp tục giải x

Làm thử từ 1 đến 7

Bài 1: 

\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x^2+2x-3}+\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{\left(x-1\right)\left(x+3\right)}+\sqrt{x-2}\)(\(ĐKXĐ:x\ge2\))

\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}\right)^2=\left(\sqrt{\left(x-1\right)\left(x+3\right)}+\sqrt{x-2}\right)^2\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)+x+3=\left(x-1\right)\left(x+3\right)+x-2\)

\(\Leftrightarrow x=2\)(nhận)

Vậy \(S=\left\{2\right\}\)

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Bài 2: 

\(\sqrt{2x^2-2x}=x^4-2x^3+x\)

\(\Leftrightarrow\sqrt{2x\left(x-1\right)}=x^4-2x^3+x\)(\(ĐKXĐ:\orbr{\begin{cases}x\le0\\x\ge1\end{cases}}\))

\(\Leftrightarrow\sqrt{2x\left(x-1\right)}=x\left(x-1\right)\left(x^2-x-1\right)\)

\(\Leftrightarrow\sqrt{x\left(x-1\right)}\left[\left(x^2-x-1\right)\sqrt{x\left(x-1\right)}-\sqrt{2}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x\left(x-1\right)}=0\left(1\right)\\\left(x^2-x-1\right)\sqrt{x\left(x-1\right)}=\sqrt{2}\left(2\right)\end{cases}}\)

Từ: \(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x=0\left(n\right)\\x=1\left(n\right)\end{cases}}\)

Từ: \(\left(2\right)\Leftrightarrow\left(x^2-x-1\right)x\left(x-1\right)=2\)

Đặt \(t=x^2-x\)(Cái này đặt hay k đặt cx đc)

\(\left(2\right)\Leftrightarrow\left(t-1\right)t=2\)

\(\Leftrightarrow t^2-t-2=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)(do \(x^2-x+1>0\forall x\))

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(n\right)\\x=-1\left(n\right)\end{cases}}\)

Vậy \(S=\left\{-1;0;1;2\right\}\)

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Bài 3:  

\(\sqrt{6x-x^2}+2x^2-12x+15=0\)(1)

\(\Leftrightarrow\sqrt{x\left(6-x\right)}-2x\left(6-x\right)+15=0\)\(\left(ĐKXĐ:0\le x\le6\right)\)

Đặt: \(t=\sqrt{x\left(6-x\right)}\left(t\ge0\right)\)

\(\left(1\right)\Leftrightarrow-2t^2+t+15=0\)

\(\Delta=1^2-4\left(-2\right)\cdot15=121>0\)

Suy ra pt có 2 nghiệm pb:

\(t_1=\frac{-1+\sqrt{\Delta}}{2\left(-2\right)}=\frac{-1+11}{-4}=\frac{-5}{2}\)(loại)

\(t_2=\frac{-1-\sqrt{\Delta}}{2\left(-2\right)}=\frac{-1-11}{-4}=3\)(nhận)

\(\Rightarrow\sqrt{x\left(6-x\right)}=3\)

\(\Leftrightarrow-x^2+6x-9=0\)

\(\Leftrightarrow-\left(x-3\right)^2=0\)

\(\Leftrightarrow x=3\)(nhận)

Vậy: \(S=\left\{3\right\}\)

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Bài 4:

\(x^2+2x+7=\left(x+3\right)\sqrt{x^2+5}\)

\(\Leftrightarrow\left(x^2+2x+7\right)^2=\left(x+3\right)^2\left(x^2+5\right)\)

\(\Leftrightarrow\left(x^2+2x+7\right)^2=\left(x^2+6x+9\right)\left(x^2+5\right)\)

\(\Leftrightarrow2x^3-4x^2+2x-4=0\)

\(\Leftrightarrow2\left(x^2+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

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Bài 5: Tương tự câu 4

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Bài 6: \(\sqrt{x^2+2x+2\sqrt{x^2+2x-1}}+2x^2+4x-4=0\)\(\left(ĐKXĐ:\orbr{\begin{cases}-1-\sqrt{3}\le x\le-1-\sqrt{2}\\\sqrt{2}-1\le x\le\sqrt{3}-1\end{cases}}\right)\)
\(\Leftrightarrow\sqrt{\left(x^2+2x-1\right)+2\sqrt{x^2+2x-1}+1}+2x^2+4x-4=0\)

\(\Leftrightarrow\sqrt{x^2+2x-1}+1+2x^2+4x-4=0\)

\(\Leftrightarrow\sqrt{x^2+2x-1}+2\left(x^2+2x-1\right)-1=0\)

 Đặt \(t=\sqrt{x^2+2x-1}\left(t\ge0\right)\)

\(PT\Leftrightarrow2t^2+t-1=0\)

\(\Leftrightarrow\left(t+1\right)\left(2t-1\right)=0\)

\(\Leftrightarrow2t-1=0\)(do t+1>0)

\(\Leftrightarrow2\sqrt{x^2+2x-1}=1\)

\(\Leftrightarrow4x^2+8x-5=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\left(n\right)\\x=\frac{1}{2}\left(n\right)\end{cases}}\)

Vậy \(S=\left\{-\frac{5}{2};\frac{1}{2}\right\}\)

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Bài 7: \(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)ĐKXĐ làm biếng tìm :D

\(\Leftrightarrow\sqrt{3x^2-7x+3}+\sqrt{x^2-3x+4}=\sqrt{3x^2-5x-1}+\sqrt{x^2-2}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(3x-1\right)+1}+\sqrt{\left(x-2\right)\left(x-1\right)+2}=\sqrt{\left(3x+1\right)\left(x-2\right)+1}+\sqrt{\left(x-2\right)\left(x+2\right)+2}\)

Tới đây k biết làm

.....

Thăm blog em nhé: tthnew's blog

Cách khác:

Bài 1:\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x^2+2x-3}+\sqrt{x-2}\)

ĐKXĐ: \(x\ge2\).

Nếu x = 2, thỏa mãn.

Nếu x > 2: Đặt \(f\left(x\right)=VT-VP\)

\(f\left(x\right)=\frac{-5x+5}{\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x+4\right)\left(x-2\right)+5}}+\frac{5}{\sqrt{x+3}+\sqrt{x-2}}\)

\(< -\sqrt{5}+\sqrt{5}=0\)

Vậy x = 2.

\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)(*)

Vì VT > 0 \(\therefore VP>0\Rightarrow x>\frac{5}{3}\).

Nhận xét x = 2 là một nghiệm. Xét x khác 2.

\(PT\Leftrightarrow\left(\sqrt{x^2+12}-x\right)+\left(x-\sqrt{x^2+5}\right)=3x-5\)

\(\Leftrightarrow\frac{12}{\sqrt{x^2+12}+x}+\frac{5}{x+\sqrt{x^2+5}}+5-3x=0\)

Nếu \(\frac{5}{3}< x< 2\Rightarrow VT>2>0\), vô nghiệm.

Nếu \(x>2\): (*) \(\Leftrightarrow\)\(\frac{7}{\sqrt{x^2+12}+\sqrt{x^2+5}}=3x-5\)

\(VT< 1< VP\). PT vô nghiệm.

Vậy x = 2.

Xong!