a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

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

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

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

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

$\Leftrightarrow |x-2|=|2x-3|$

\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

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

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)

d.

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

$\Leftrightarrow |2x-1|=|x-3|$

\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)

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

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)

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

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

a/ ĐKXĐ: $x\geq 1$

Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:

$a+b+2ab=6-(a^2+b^2)$

$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$

$\Leftrightarrow (a+b)^2+(a+b)-6=0$

$\Leftrightarrow (a+b-2)(a+b+3)=0$

Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$

$\Leftrightarrow a+b=2$

Mà $b^2-a^2=(x+3)-(x-1)=4$

$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$

$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$

$\Leftrightarrow x=1$ (tm)

b/

ĐKXĐ: $x\geq 1$

Đặt $\sqrt{3x-2}=a; \sqrt{x-1}=b(a,b\geq 0)$. Khi đó pt đã cho trở thành:

$a+b=a^2+b^2-6+2ab$

$\Leftrightarrow a^2+b^2+2ab-(a+b)-6=0$

$\Leftrightarrow (a+b)^2-(a+b)-6=0$

$\Leftrightarrow (a+b+2)(a+b-3)=0$

Hiển nhiên $a+b+2>0$ với mọi $a,b\geq 0$

Do đó $a+b-3=0\Leftrightarrow a+b=3$

$\Leftrightarrow b=3-a$.

Ta thấy $a^2-3b^2=1$. Thay $b=3-a$ vô thì:

$a^2-3(3-a)^2=1$

$\Leftrightarrow (a-2)(a-7)=0$

$\Leftrightarrow a=2$ hoặc $a=7$

Vì $a+b=3$ mà $a,b>0$ nên $a,b<3$. Do đó $a=2$

$\Leftrightarrow \sqrt{3x-2}=2$ 

$\Leftrightarrow x=2$ 

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?

Câu 1:ĐK \(x\ge\frac{1}{2}\)

\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)

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

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

<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)

<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)

Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)

=> \(x=1\)(TM ĐKXĐ)

Vậy x=1

câu 2 ĐK \(x\ge1\)

\(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}=0\)

<=> \(\left(5x+8\right)\left(\sqrt{2x-1}-1\right)+7x\left(\sqrt{x+3}-2\right)+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)

<=>\(\left(5x+8\right).\frac{2x-2}{\sqrt{2x-1}+1}+7x.\frac{x+3-4}{\sqrt{x+3}+2}+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)

<=> \(\sqrt{x-1}\left(\frac{2\left(5x+8\right)\sqrt{x-1}}{\sqrt{2x-1}+1}+\frac{7x\sqrt{x-1}}{\sqrt{x+3}+2}+\left(x+26\right)+10\sqrt{x-1}\right)=0\)

Với \(x\ge1\)thì cái trong ngoặc >0

=> \(x=1\)

Vậy x=1

1) Ta có: \(\left|x^2-4x-5\right|=x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x-1\left(\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\right)\\-x^2+4x+5=x-1\left(-1< x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5-x+1=0\\-x^2+4x+5-x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=0\\-x^2+3x+6=0\end{matrix}\right.\Leftrightarrow x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{41}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{\sqrt{41}}{2}\\x-\dfrac{5}{2}=-\dfrac{\sqrt{41}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}+5}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}+5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{41}+5}{2}\right\}\)