2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

giải bằng cách đặt ẩn nha các bn

Đặt x/(x^2-3x+3) = t ta được 

\(3t-2t=1\Leftrightarrow t=1\)

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

\(\Leftrightarrow\left(x-2\right)^2-1=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\Leftrightarrow x=3;x=1\)

\(1,\\ a,=x^2+6x+9-x^2-6x=9\\ b,=3x-1+6x-9x^2+x-10=-9x^2+10x-11\\ 2,\\ a,=4xy\left(x^2-2xy+y^2\right)=4xy\left(x-y\right)^2\)

ĐKXĐ là x\(\ge\frac{2}{3}\)

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

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

a) 2x³+5x²-3x=0

<=> 2x³+6x²-x²-3x=0

<=> 2x²(x+3)-x(x+3)=0

<=> (x+3)(2x²-x)=0

<=> (x+3)x(2x-1)=0

\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

c) x³+1=x(x+1)

<=> (x+1)(x²+1-x)-x(x+1)=0

<=> (x+1)(x²-2x+1)=0

<=> (x+1)(x-1)²=0

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy ...

Thanks !

\(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)

\(\Rightarrow3x^2-3x+2x-2-3\left(x^2-2x+x-2\right)=4\)

\(\Rightarrow3x^2-x-2-3x^2+3x+6=4\)

\(\Rightarrow2x+4=4\)

\(\Rightarrow2x=0\Leftrightarrow x=0\)

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