Đáp án:
Giải thích các bước giải:
`a)`
`4x(x+1)=8(x+1)`
`-> 4x(x+1) - 8(x+1) = 0`
`-> 4x^2-4x-8=0`
`-> 4(x^2-x-2) = 0`
`-> 4[(x^2+x)+(-2x-2)]=0`
`-> 4[x(x+1)-2(x+1)] =0`
`-> 4(x+1)(x-2) = 0`
`->`\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy `x \in {2,-1}`
`b)`
`x(x-1)-2(1-x)=0`
`-> x^2 + x - 2 = 0`
`-> (x^2-x)+(2x-2)=0`
`-> x(x-1)+2(x-1) = 0`
`-> (x-1)(x+2)=0`
`->`\(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy `x \in {1,-2}`
`c)`
`2x(x-2)-(2-x)^2=0`
`-> 2x^2 - 4x - 4 + 4x - x^2 = 0`
`-> x^2 - 4 = 0`
`-> x^2 - 2^2 = 0`
`-> (x-2)(x+2) = 0`
`->`\(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `x \in {-2,2}`
`d)`
`(x-3)^3 + 3 - x = 0`
`-> x^3 - 9x^2 + 26x - 24 = 0`
`-> (x-4)(x^2-5x+6)=0`
`-> (x-4)[(x^2-3x)+(-2x-6)]=0`
`-> (x-4)[x(x-3)-2(x-3)] = 0`
`-> (x-4)(x-3)(x-2) = 0`
`->`\(\left[ \begin{array}{l}x-2=0\\x-3=0\\x-4=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=2\\x=3\\x=4\end{array} \right.\)
Vậy `x \in {2,3,4}`