Question

Giải pt: \(\left(1-\sqrt{1-x}\right)\sqrt[3]{2-x}=x\)

Answer 1

ĐK \(x\le1\)

PT

<=> \(\left(1-1+x\right)\sqrt[3]{2-x}=x\left(1+\sqrt{1-x}\right)\)

<=> \(x\left(\sqrt{1-x}+1-\sqrt[3]{2-x}\right)=0\)

<=> \(\left[{}\begin{matrix}x=0\left(tmĐK\right)\\\sqrt{1-x}+1=\sqrt[3]{2-x}\left(2\right)\end{matrix}\right.\)

(2) Đặt \(\sqrt{1-x}=a;\sqrt[3]{2-x}=b\left(a\ge0\right)\)

=> \(\left\{{}\begin{matrix}a+1=b\\a^2-b^3=-1\end{matrix}\right.\)

=> \(\left(b-1\right)^2-b^3=-1\)

=> \(b=1\)=> \(x=1\)(tm ĐK)

Vậy x=1;x=0