Đáp án:
Giải thích các bước giải:
`(x+1)(x+2)=1`
`⇔ x^2+2x+x+2=1`
`⇔ x^2+3x+2-1=0`
`⇔ x^2+3x+1=0`
`⇔ x^2+2 . 3/2 x+9/4-5/4=0`
`⇔ (x+3/2)^2-(\frac{\sqrt{5}}{2})^2=0`
`⇔ (x+3/2-\frac{\sqrt{5}}{2})(x+3/2+\frac{\sqrt{5}}{2})=0`
`⇔` \(\left[ \begin{array}{l}x+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}=0\\x+\dfrac{3}{2}+\frac{\sqrt{5}}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{-3-\sqrt{5}}{2}\\x=\dfrac{-3+\sqrt{5}}{2}\end{array} \right.\)
Vậy `S={\frac{-3-\sqrt{5}}{2};\frac{-3+\sqrt{5}}{2}}`
Sửa đề:
`(x+1)(x-2)=1`
`⇔ x^2-2x+x-2=1`
`⇔ x^2-x-2-1=0`
`⇔ x^2-x-3=0`
`⇔ x^2-2 . 1/2 x+1/4-13/4=0`
`⇔ (x-1/2)^2-(\frac{\sqrt{13}}{2})^2=0`
`⇔ (x-1/2-\frac{\sqrt{13}}{2})(x-1/2+\frac{\sqrt{13}}{2})=0`
`⇔` \(\left[ \begin{array}{l}x-\dfrac{1}{2}-\dfrac{\sqrt{13}}{2}=0\\x-\dfrac{1}{2}+\dfrac{\sqrt{13}}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{1+\sqrt{13}}{2}\\x=\dfrac{1-\sqrt{13}}{2}\end{array} \right.\)
Vậy `S={\frac{1-\sqrt{13}}{2};\frac{1+\sqrt{13}}{2}}`
