Question

Answer 1

`\sqrt{(1+x)(2-x)}=1+2x-2x^2`

`<=>\sqrt{2+x-x^2}=2(2+x-x^2)-3`

Đặt `\sqrt{2+x-x^2}=t` `(t >= 0)` khi đó ptr có dạng:

      `t=2t^2-3`

`<=>2t^2-t-3=0`

`<=>2t^2-3t+2t-3=0`

`<=>t(2t-3)+(2t-3)=0`

`<=>(2t-3)(t+1)=0`

`<=>` $\left[\begin{matrix} t=\dfrac{3}{2}\text{ (t/m)}\\ t=-1\text{ (ko t/m)}\end{matrix}\right.$

    `@t=3/2=>\sqrt{2+x-x^2}=3/2`

             `<=>2+x-x^2=9/4`

             `<=>x^2-x+1/4=0`

             `<=>(x-1/2)^2=0`

            `<=>x=1/2`

Vậy `S={1/2}`

Answer 2

help ;-;