đặt x^2-x=t

pt=>3t^2-2t=0=>t(3t-2)=0

tự giải tiếp........=)

Kết quả là bao nhiêu v? 

\(\dfrac{1-2x}{x+2}\in Z\Leftrightarrow\dfrac{-2\left(x+2\right)+5}{x+2}\in Z\Leftrightarrow-2+\dfrac{5}{x+2}\in Z\\ \Leftrightarrow x+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow x\in\left\{-7;-3;-1;3\right\}\\ \Leftrightarrow A=\left\{-7;-3;-1;3\right\}\)

A={-1;-3;3;-7}

1: A=[-3;6)

C={1;3}

2: B\(\cap\)C={1}

A\B=[-3;-1)

Với `k=0` :

\(x=2.0.\left(0+2\right)=0\left(TM\right)\)

Với k = 1 :

\(x=2.1.\left(1+2\right)=6\left(TM\right)\)

Tương tự với `k=2,3`

\(=>Q=\left\{0;6;16;30\right\}\)

Giải phương tình: \(x+\sqrt{2x-1}=2\left(x-3\right)^2\)

Điều kiện: \(x\ge\dfrac{1}{2}\)

\(PT\Leftrightarrow\sqrt{2x-1}-3=2x^2-13x+15\\ \Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}-3}=\left(x-5\right)\left(2x-3\right)\\ \Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}-2x+3\right)=0\\ \Leftrightarrow\begin{matrix}x=5\\\dfrac{2}{\sqrt{2x-1}+3}=2x-3\left(1\right)\end{matrix}\)

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

Đặt \(t=\sqrt{2x-1},t>0\) phương trình trở thành \(\left(t^2-2\right)\left(t+3\right)=2\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}t=-2\left(L\right)\\t=\dfrac{-1-\sqrt{17}}{2}\\t=\dfrac{-1+\sqrt{17}}{2}\end{matrix}\right.\left(L\right)\)

Với \(t=\dfrac{-1+\sqrt{17}}{2}\) ta có \(\sqrt{2x-1}=\dfrac{-1+\sqrt{17}}{2}\)

\(\Leftrightarrow2x-1=\dfrac{9-\sqrt{17}}{2}\)

\(\Leftrightarrow x=\dfrac{11-\sqrt{17}}{4}\)

Vậy \(E=\left\{5;\dfrac{11-\sqrt{17}}{4}\right\}\)