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

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

Cái này Liên ợp thần chưởng thôi !

ĐK: \(\frac{10}{3}\ge x\ge\frac{6}{5}\)ta có pt 

<=>\(2x^2-4x+3x-6=\sqrt{5x-6}-2+\sqrt{10-3x}-2\)

<=>\(2x\left(x-2\right)+3\left(x-2\right)=\frac{5\left(x-2\right)}{\sqrt{5x-6}+2}+\frac{3\left(2-x\right)}{\sqrt{10-3x}+2}\)

<=>\(\left(x-2\right)\left(2x+3+\frac{3}{\sqrt{10-3x}+2}-\frac{5}{\sqrt{5x-6}+2}\right)=0\) (1)

Vì \(\sqrt{5x-6}+2\ge2\Rightarrow\frac{-5}{\sqrt{5x-6}+2}\ge-\frac{5}{2}\)

Mà \(x\ge\frac{6}{5}\Rightarrow2x+3-\frac{5}{\sqrt{5x-6}+2}+\frac{3}{\sqrt{10-3x}+2}>0\)

Nên pt(1) <=> x=2 (thỏa mãn ĐK)

vậy ...

^_^

\(\sqrt{5x-6}+\sqrt{10-3x}=2x^2-x-2\)

\(\Leftrightarrow\sqrt{5x-6}-2x^2+x+\sqrt{10-3x}+2=0\)

\(\Leftrightarrow x=2\)

\(\sqrt{5x-6}+\sqrt{10-3x}=2x^2-x-2\)

\(\Leftrightarrow\sqrt{5x-6}-2x^2+x+\sqrt{10-3x}+2=0\)

\(\Leftrightarrow x=2\)

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

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

ĐK:\(x\ge3\)

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

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

\(\Leftrightarrow\frac{x^2-5x+4}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x-4}{\sqrt{x+1}+\sqrt{5}}=\frac{x-4}{\sqrt{x-2}+\sqrt{2}}+\frac{x^2-2x-8}{\sqrt{x^2-2x-3}+\sqrt{5}}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-4\right)}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x-4}{\sqrt{x+1}+\sqrt{5}}-\frac{x-4}{\sqrt{x-2}+\sqrt{2}}-\frac{\left(x-4\right)\left(x+2\right)}{\left(x+2\right)\sqrt{x^2-2x-3}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-4\right)\left(\frac{x-1}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{1}{\sqrt{x+1}+\sqrt{5}}-\frac{1}{\sqrt{x-2}+\sqrt{2}}-\frac{x+2}{\left(x+2\right)\sqrt{x^2-2x-3}+\sqrt{5}}\right)=0\)

Suy ra x-4=0 =>x=4

đề bài có đúng không

chắc chắn đúng

ĐK: \(x\ge3\)

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

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

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

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

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

Đến đây dễ rồi bn tự làm tiếp nhé