\(a,x\left(x-1\right)+x-2=0\\ \Leftrightarrow x^2+2x+x-2=0\\ \Leftrightarrow x^2-x-2=0\)
\(x=\dfrac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4.1.\left(-2\right)}}{2.1}\\ x=\dfrac{1\pm\sqrt{1+8}}{2}\\ x=\dfrac{1\pm3}{2}\)
Xét 2 trường hợp:
*\(x=\dfrac{1+3}{2}\\ =>x=2\)
\(x=\dfrac{1-3}{2}\\ =>x=-1\)
Vậy.........................
1.Tìm x:
a, x(x-2)+ x-2 = 0
⇔ (x+1)(x-2)= 0
⇔ \(\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy x = -1 hoặc x = 2
b, (2x-1)2 - (x+2)2 = 0
⇔ (2x-1)2 = (x+2)2
⇔ 2x-1 = x+2
⇔ 2x - x = 2 +1
⇒ x = 3
Vậy x = 3
c, x3- x = 0
⇔ x(x2 - 1) = 0
⇔ \(\left\{{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy x = 0 hoặc x = 1
\(a.x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(b.\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=\left(x+2\right)^2\Leftrightarrow2x-1=x+2\)
\(\Leftrightarrow2x-x=2+1\Leftrightarrow x=3\)
\(c.x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
