Đáp án:
`a)x=-1`
`b)x∈{0;1}`
`c)x∈{0;1;-1/2}`
`d)x=2`
Giải thích các bước giải:
`a)x²(x+1)+x+1=0`
`⇔x²(x+1)+(x+1)=0`
`⇔(x+1)(x²+1)=0`
Ta có:`x²≥0∀x`
`⇒x²+1≥1>0∀x`
`⇒` vô nghiệm
`⇔x+1=0`
`⇔x=-1`
Vậy `x=-1`
`b)x²-x=-2x²+2x`
`⇔x²-x+2x²-2x=0`
`⇔(x²+2x²)-(x+2x)=0`
`⇔3x²-3x=0`
`⇔3x(x-1)=0`
`⇔`$\left[\begin{matrix} 3x=0\\ x-1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0\\ x=1\end{matrix}\right.$
Vậy `x∈{0;1}`
`c)2x²(x-1)+x²=x`
`⇔2x²(x-1)+x²-x=0`
`⇔2x²(x-1)+(x²-x)=0`
`⇔2x²(x-1)+x(x-1)=0`
`⇔(x-1)(2x²+x)=0`
`⇔x(x-1)(2x+1)=0`
`⇔`$\left[\begin{matrix} x=0\\ x-1=0\\2x+1=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0\\ x=1\\x=-\dfrac{1}{2}\end{matrix}\right.$
Vậy `x∈{0;1;-1/2}`
`d)(x-2)(x²+4)=x²-2x`
`⇔(x-2)(x²+4)=x(x-2)`
`⇔(x-2)(x²+4)-x(x-2)=0`
`⇔(x-2)(x²+4-x)=0`
`⇔(x-2)(x²-x+4)=0`
`⇔(x-2)(x²-x+1/4+15/4)=0`
`⇔(x-2)[(x²-x+1/4)+15/4]=0`
`⇔(x-2){[x²-2.x. 1/2+(1/2)^2]+15/4}=0`
`⇔(x-2)[(x-1/2)^2+15/4]=0`
Ta có:`(x-1/2)^2≥0∀x`
`⇒(x-1/2)^2+15/4≥15/4>0∀x`
`⇒` vô nghiệm
`⇔x-2=0`
`⇔x=2`
Vậy `x=2`
