Đáp án+Giải thích các bước giải:
`a)` `(x^2-4)-(2-x)(2x-3)=0`
`<=>(x-2)(x+2)+(x-2)(2x-3)=0`
`<=>(x-2)(3x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={1/3;2}`
`b)` `(2x-1)^3=2(1-4x^2)`
`<=>8x^3-12x^2+6x-1=-8x^2+2`
`<=>8x^3-4x^2+6x-3=0`
`<=>(2x-1)(4x^2+3)=0`
Ta có `4x^2>=0=>4x^2+3>=3`
`<=>2x-1=0`
`<=>x=1/2`
Vậy `S={1/2}`
`c)` `(x-1)^2+2(1-x^2)=0`
`<=>-x^2-2x+3=0`
`<=>(x+3)(-x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=-3\\x=1\end{array} \right.\)
Vậy `S={1;-3}`
`d)` `(x+1/x)^2+2(x+1/x)-8=0`
`<=>(x^4+2x^3-6x^2+2x+1)/(x^2)=0`
`<=>x^4+2x^3-6x^2+2x+1=0`
`<=>(x-1)^2(x^2+4x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{array} \right.\)
Vậy `S={1;-2+\sqrt{3};x=-2-\sqrt{3}}`
