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a: x(x+5)^2=x+5

=>(x+5)(x^2+5x-1)=0

=>x+5=0 hoặc x^2+5x-1=0

=>\(x\in\left\{-5;\dfrac{-5+\sqrt{29}}{2};\dfrac{-5-\sqrt{29}}{2}\right\}\)

b: x(x-2)=(x-2)

=>(x-2)(x-1)=0

=>x=2 hoặc x=1

`x^2 +3(x-1/2)=x^2+3`

`=>x^2+3x-3/2 =x^2+3`

`=> x^2 +3x-x^2=3+3/2`

`=> 3x=6/2+3/2`

`=>3x= 9/2`

`=>x= 9/2 : 3`

`=>x= 9/6= 3/2`

Vậy `x=3/2`

\(x^2+3\left(x-\dfrac{1}{2}\right)=x^2+3\\ \Rightarrow x^2-x^2+3x-\dfrac{3}{2}=3\\ \Rightarrow3x=\dfrac{3}{2}+3\\ \Rightarrow3x=\dfrac{9}{2}\\ \Rightarrow x=\dfrac{9}{2}:3\\ \Rightarrow x=\dfrac{3}{2}\)

Vậy \(x\in\left\{\dfrac{3}{2}\right\}\)

=>x^2+x-x+2=11

=>x^2=9

=>x=3 hoặc x=-3

\(x\left(x+1\right)-\left(x-2\right)=11\)

\(\Rightarrow x^2+x-x+2=11\)

\(\Rightarrow x^2+2=11\)

\(\Rightarrow x^2=11-2\)

\(\Rightarrow x^2=9\)

\(\Rightarrow x^2=3^2\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Bài 1: 

a: x/-2=-18/x

=>x²=36

=>x=6 hoặc x=-6

b: x/2+x/5=17/10

=>7/10x=17/10

hay x=17/7

\(x-\dfrac{2}{3}.\left(x+9\right)=1\)

\(\Rightarrow x-\dfrac{2}{3}x-6=1\)

\(\Rightarrow\dfrac{1}{3}x=7\)

\(\Rightarrow x=21\)

\(x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}x-6=1\)

\(\dfrac{1}{3}x-6=1\)

\(\dfrac{1}{3}x=1+6\)

\(\dfrac{1}{3}x=7\)

\(x=7:\dfrac{1}{3}\)

\(x=21\)

\(x^2\left(x-5\right)+x-5=0\)

\(\Rightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)

\(x^2\left(x-5\right)+x-5=0\)

\(\Leftrightarrow x-5=0\)

hay x=5

=>4,5x=27

hay x=6