a: ĐKXĐ: 2<=x<=10

\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)

=>\(\sqrt{x-2}-2+\sqrt{10-x}-2=x^2-12x+36\)

=>\(\frac{x-2-4}{\sqrt{x-2}+2}+\frac{10-x-4}{\sqrt{10-x}+2}=\left(x-6\right)^2\)

=>\(\left(x-6\right)\cdot\left(\frac{1}{\sqrt{x-2}+2}-\frac{1}{\sqrt{10-x}+2}-x+6\right)=0\)

=>x-6=0

=>x=6(nhận)

b: ĐKXĐ: 5/3<=x<=7/3

\(\sqrt{3x-5}+\sqrt{7-3x}=5x^2-20x+22\)

=>\(\sqrt{3x-5}-1+\sqrt{7-3x}-1=5x^2-20x+20\)

=>\(\frac{3x-5-1}{\sqrt{3x-5}+1}+\frac{7-3x-1}{\sqrt{7-3x}+1}=5\left(x^2-4x+4\right)\)

=>\(\frac{3\left(x-2\right)}{\sqrt{3x-5}+1}+\frac{-3\left(x-2\right)}{\sqrt{7-3x}+1}-5\left(x-2\right)^2=0\)

=>(x-2)\(\left\lbrack\frac{3}{\sqrt{3x-5}+1}-\frac{3}{\sqrt{7-3x}+1}-5\left(x-2\right)\right\rbrack=0\)

=>x-2=0

=>x=2(nhận)

c: ĐKXĐ: x∈R

\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)

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

=>|x-2|+|x-3|=1

mà \(\left|x-2\right|+\left|x-3\right|\ge\left|x-2-x+3\right|=1\forall x\)

nên dấu '=' xảy ra khi (x-2)(x-3)<=0

=>2<=x<=3

a, ĐK: ​\(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)​

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

\(\Leftrightarrow x^2+2x-1+3\left(x+1\right)-4\sqrt{\left(x+1\right)\left(x^2+2x-1\right)}=0\)

TH1: \(x\ge-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=\sqrt{x+1}\\\sqrt{x^2+2x-1}=3\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

TH2: \(x< -1\)

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

\(\Leftrightarrow...\)

Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)

\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)

đặt\(x^2+x+1=t\left(t>0\right)\)

\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)

bình phương 2 vế pt trở thành:

\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)

\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m

vậy pt vô nghiệm

a/ ĐKXĐ: ...

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)

\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)

\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))

\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)

\(\Leftrightarrow11a^2+6a-25=0\)

Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó

b/

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)

\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)

\(\Leftrightarrow\sqrt{a^2+3a}=2\)

\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+1=1\)

ĐKXĐ:...

\(\sqrt{3x^2-5x-1}-\sqrt{3x^2-7x+9}+\sqrt{x^2-2}-\sqrt{x^2-3x+13}=0\)

\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3\left(x-5\right)}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}\right)=0\)

\(\Leftrightarrow x-5=0\) (ngoặc to phía sau luôn dương)

\(\Rightarrow x=5\)

Akai Haruma @Nguyễn Việt Lâm

a:

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

=>|x-3|=3

=>x-3=3 hoặc x-3=-3

=>x=0 hoặc x=6

b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)

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

=>\(\left|\sqrt{x-1}+1\right|=2\)

=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)

=>x-1=1

=>x=2

c:

ĐKXĐ: x>4/5

PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

d: ĐKXĐ: x-4>=0 và x+1>=0

=>x>=4

PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)

=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)

=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)

=>\(\sqrt{x^2-3x-4}=14-x\)

=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196

=>x<=14 và -3x-4=-28x+196

=>x<=14 và 25x=200

=>x=8(nhận)

a) \(\sqrt{x^2-6x+9}=3\)

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

\(\Leftrightarrow\left|x-3\right|=3 \)

TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)

Pt trở thành:

\(x-3=3\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=3+3\)

\(\Leftrightarrow x=6\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)

Pt trở thành:

\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=-3+3\)

\(\Leftrightarrow x=0\left(tm\right)\)

b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))

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

\(\Leftrightarrow2\sqrt{x-1}=4-x\)

\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)

\(\Leftrightarrow4x-4=16-8x+x^2\)

\(\Leftrightarrow x^2-12x+20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))

\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)

\(\Leftrightarrow5x-4=4x+8\)

\(\Leftrightarrow x=12\left(tm\right)\)