a/ Giải rồi

b/ ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

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

Pt trở thành:

\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)

\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)

\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)

\(\Leftrightarrow...\)

e/ ĐKXD: \(x>0\)

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

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)

\(\Rightarrow t^2=x+\frac{1}{4x}+1\)

Pt trở thành:

\(5t=2\left(t^2-1\right)+4\)

\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)

\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)

Câu 6:

ĐK: $x\geq 1$

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

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

$\Leftrightarrow |\sqrt{x-1}-1|=\sqrt{x-1}+1$

Nếu $\sqrt{x-1}-1\geq 0$ thì PT trở thành:

$\sqrt{x-1}-1=\sqrt{x-1}+1\Leftrightarrow 2=0$ (vô lý)

Nếu $\sqrt{x-1}-1< 0$ (tương đương với $1\leq x< 2$ thì PT trở thành:

$1-\sqrt{x-1}=\sqrt{x-1}+1$

$\Leftrightarrow \sqrt{x-1}=0\Rightarrow x=1$ (thỏa mãn)

Vậy PT có nghiệm $x=1$

Câu 5:

ĐK: $x\geq 1$

PT $\Leftrightarrow \sqrt{(x-1)-4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1$

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

$\Leftrightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$

Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:

$|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=|\sqrt{x-1}-2|+|3-\sqrt{x-1}|\geq |\sqrt{x-1}-2+3-\sqrt{x-1}|=1$

Dấu "=" xảy ra khi $(\sqrt{x-1}-2)(3-\sqrt{x-1})\geq 0$

$\Leftrightarrow 3\geq \sqrt{x-1}\geq 2$

$\Leftrightarrow 10\geq x\geq 5$. Kết hợp ĐKXĐ ta thấy những giá trị $x$ thỏa mãn $10\geq x\geq 5$ là nghiệm của pt.

a) \(\sqrt{4x}=10\) (ĐKXĐ: 4x>=0 <=> x>=0)

\(\Leftrightarrow4x=100\)

\(\Leftrightarrow x=25\)

\(S=\left\{25\right\}\)

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

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=8\)

\(\Leftrightarrow x-1=8\)

\(\Leftrightarrow x=9\)

\(S=\left\{9\right\}\)

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

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

\(\Leftrightarrow x-3=1-3x\) hoặc \(\Leftrightarrow x-3=-1+3x\)

\(\Leftrightarrow x+3x=1+3\) \(\Leftrightarrow x-3x=-1+3\)

\(\Leftrightarrow4x=4\) \(\Leftrightarrow-2x=2\)

\(\Leftrightarrow x=1\) \(\Leftrightarrow x=-1\)

\(S=\left\{1;-1\right\}\)

d) \(\sqrt{2x-5}=x-2\)

\(\Leftrightarrow2x-5=x^2-4x+4\)

\(\Leftrightarrow-x^2+2x+4x-5-4=0\)

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

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

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

\(S=\left\{3\right\}\)

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

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

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

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

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

\(\Leftrightarrow x=0\) hoặc \(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

\(S=\left\{0;3\right\}\)

g) \(\sqrt{x^2-9}-\sqrt{x-3}=0\) ( ĐKXĐ: x-3>=0 <=> x>=3)

\(\Leftrightarrow\sqrt{x^2-9}=\sqrt{x-3}\)

\(\Leftrightarrow x^2-9=x-3\)

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

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)

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

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

\(\Leftrightarrow x+2=0\) hoặc \(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=-2\) \(\Leftrightarrow x=3\)

\(S=\left\{-2;3\right\}\)

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

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

\(\Leftrightarrow x-2+x-3-1=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\)

\(S=\left\{3\right\}\)

i) \(\sqrt{\frac{2x-3}{x-1}}=2\)

\(\Leftrightarrow\frac{2x-3}{x-1}=4\)

\(\Leftrightarrow4\left(x-1\right)=2x-3\)

\(\Leftrightarrow4x-4-2x+3=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(S=\left\{\frac{1}{2}\right\}\)

l) \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)

\(\Leftrightarrow x+y-4\sqrt{x}+12-6\sqrt{y-1}=0\)

\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-1-6\sqrt{y-1}+9\right)=0\)

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

\(\Leftrightarrow\sqrt{x}-2=0\) hoặc \(\Leftrightarrow\sqrt{y-1}-3=0\)

\(\Leftrightarrow\sqrt{x}=2\) \(\Leftrightarrow\sqrt{y-1}=3\)

\(\Leftrightarrow x=4\) \(\Leftrightarrow y-1=9\)

\(\Leftrightarrow y=10\)

KẾT luận : ..............

Tới đây nhé, nếu mai chưa ai giải thì mình giải hộ cho

CHÚC BẠN HỌC TỐT!

m) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)

<=> \(\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)

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

<=>\(\sqrt{x-1}+2+\sqrt{x-1}+3=5\)

<=> \(2\sqrt{x-1}=0\)

<=> \(\sqrt{x-1}=0\) <=>x=1

Vậy \(S=\left\{1\right\}\)

n) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) (*) ( đk \(x\ge\frac{1}{2}\))

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

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

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

<=> x+\(\left|x-1\right|=2\)(1)

TH1: \(\frac{1}{2}\le x\le1\)

Từ (1) => x+1-x=2

<=> 1=2(vô lý)

TH2: x>1

Từ (1)=> x+x-1=2

<=> 2x=3<=> \(x=\frac{2}{3}\)(tm pt (*))

Vậy \(S=\left\{\frac{2}{3}\right\}\)

p) \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\) (*) (đk :\(x\ge2\))

Đặt \(\left\{{}\begin{matrix}x-2=a\left(a\ge0\right)\\x+1=b\left(b\ge0\right)\end{matrix}\right.\) =>a+b=2x-1

Có \(\sqrt{a+b}+\sqrt{a}=\sqrt{b}\)

<=> \(\sqrt{a+b}=\sqrt{b}-\sqrt{a}\)

<=> \(a+b=b-2\sqrt{ab}+a\)

<=> 0=\(-2\sqrt{ab}\)

=> \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) => x=2 (vì x=-1 không thỏa mãn pt(*))

Vậy \(S=\left\{2\right\}\)

q) \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)(*) (đk : \(7\le x\le9\))

Với a,b\(\ge0\) có: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\)(tự cm nha) .Dấu "=" xảy ra <=> a=b

Áp dụng bđt trên có:

\(\sqrt{x-7}+\sqrt{9-x}\le2\sqrt{\frac{x-7+9-x}{2}}=2\sqrt{\frac{2}{2}}=2\) (1)

Có x²-16x+66=(x²-16x+64)+2=(x-8)²+2 \(\ge2\) với mọi x (2)

Từ (1),(2) .Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-7=9-x\\x-8=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}2x=16\\x=8\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=8\\x=8\end{matrix}\right.\)<=> x=8( tm pt (*))

Vậy \(S=\left\{8\right\}\)

7.

ĐKXĐ: ...

\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow10ab=3\left(a^2+b^2\right)\)

\(\Leftrightarrow3a^2-10ab+3b^2=0\)

\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)

6.

ĐKXĐ: ...

\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2a^2+2b^2=3ab\)

\(\Leftrightarrow2a^2-3ab+2b^2=0\)

Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)

I) xd mọi x

\(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)

\(\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-5\right)^2}=9=>\left|x-4\right|+\left|x-5\right|=9\)

\(\left[{}\begin{matrix}x< 4\Rightarrow4-x+5-x=>x=0\left(n\right)\\4\le x< 5\Rightarrow x-4+5-x=9\left(vn\right)\\x\ge5\Rightarrow x-4+x-5=9\Rightarrow x=9\left(n\right)\\\end{matrix}\right.\)

kết luận

\(\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)     ( SỬA ĐỀ)

\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)

\(|x-1-2|+|x-1-3|=1\)

\(|x-3|+|x-4|=1\)

Với  \(x\le3\)thì  PT thành  \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)

Với  \(3\le x< 4\)thì PT thành  \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)

Với  \(x\ge4\)thì PT thành  \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)

Vậy  \(3\le x\le4\)

Dấu căn của x-1 đâu bạn j eiiiii

a/ ĐK: \(x \ge -1\). Đặt \(\sqrt{x+1}=a \ge 0\)
PT: \(\Leftrightarrow6a-3a-2a=5\)
\(\Leftrightarrow a=5\)
\(\Leftrightarrow x+1=15\Leftrightarrow x=24\) (nhận)

b,c: Hai ý này đều làm theo cách bình phương hoặc đưa về phương trình chứa dấu giá trị tuyệt đối được nhé.

b) Cách 1: ĐKXĐ: Tự tìm
\(\sqrt{x^{2}-4x+4}=2\Leftrightarrow x^{2}-4x+4=4\Leftrightarrow x(x-4)=0\)
\(\Leftrightarrow x=0\) hoặc \(x=4\) cả 2 cái này đều TMĐK

Cách 2: \((\sqrt{x^2-4x+4}=2)\)
\(\Leftrightarrow \sqrt{(x-2)^2}=2\)
\(\Leftrightarrow \mid x-2\mid=2\)
Với \(x\geq 2\) thì :
\(x-2=2 \Leftrightarrow x=4\) (nhận)
Với \(x<2\) thì
\(-x-2=2\Leftrightarrow x=0\) (nhận)
Vậy \(S={0;4}\)

c) Cách 1: \(\sqrt{x^{2}-6x+9}=x-2\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x^{2}-6x+9=x^{2}-4x+4 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x=\frac{5}{2} \end{matrix}\right.\)
Nghiệm TMĐK

Cách 2: \((\sqrt{x^2-6x+9}=x-2)\)
\(\Leftrightarrow \mid x-3\mid =x-2\)
Với \(x\geq 3\) thì
\(x-3=x-2\Leftrightarrow 0x=-1\) ( vô lý)
Với \(x<3\) thì
\(-x+3=x-2\Leftrightarrow -2x=-5 \Leftrightarrow x=\frac{5}{2}\)
Vậy \(S={\frac{5}{2}}\)
d) ĐKXĐ: Tự tìm
\(\sqrt{x^{2}+4}=\sqrt{2x+3}\Leftrightarrow x^{2}+4=2x+3\Leftrightarrow x^{2}-2x+1=0\Leftrightarrow (x-1)^{2}=0\)
\(\Leftrightarrow x=1\)
e) ĐKXĐ: \(x\geq \frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow \frac{2x-3}{x-1}=4\Rightarrow 2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\)
Nghiệm không TMĐK.
Phương trình vô nghiệm.
f) ĐKXĐ: \(x\geq \frac{-15}{2}\)
\(x+\sqrt{2x+15}=0\Leftrightarrow 2x+2\sqrt{2x+15}=0\Leftrightarrow 2x+15+2\sqrt{2x+15}+1-16=0\)
\(\Leftrightarrow (\sqrt{2x+15}+1)^{2}-4^{2}=0\Leftrightarrow (\sqrt{2x+15}+5)(\sqrt{2x+15}-3)=0\)
\(\Leftrightarrow \sqrt{2x+15}-3=0\Leftrightarrow \sqrt{2x+15}=3\Leftrightarrow 2x+15=9\Leftrightarrow x=-3\) (TMĐK)

Giời, có thế cũng hok hiểu, lật sách giải ra coi :v

1)

ĐK: \(x\geq 5\)

PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)

2)

ĐK: \(x\geq -1\)

\(\sqrt{x+1}+\sqrt{x+6}=5\)

\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)

\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)

Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$

\(\Rightarrow x=3\) (thỏa mãn)

Vậy .............