Question

Giải phương trình:

a)\(\sqrt[3]{x+1}-1=x\)

b)\(\sqrt[3]{1-x}+\sqrt[3]{1+x}=2\)

c)\(\sqrt[3]{x}+\sqrt[3]{2x-3}=\sqrt[3]{12\left(x-1\right)}\)

Answer 1

\(a)\sqrt[3]{x+1}-1=x\\ \Leftrightarrow\sqrt[3]{x+1}=x+1\\ \Leftrightarrow x+1=\left(x+1\right)^3\\ \Leftrightarrow x+1-\left(x+1\right)^3=0\\ \Leftrightarrow\left(x+1\right)\left(1-\left(x+1\right)^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\1-\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)

Vậy...

Answer 2

b) Phương trình tương đương:

\(1+6\sqrt[3]{1-x^2}+1=8\\ \Leftrightarrow6\sqrt[3]{1-x^2}=6\\ \Leftrightarrow\sqrt[3]{1-x^2}=1\\ \Leftrightarrow1-x^2=1\\ \Leftrightarrow-x^2=0\\ \Leftrightarrow x=0\)

Vậy \(x=0\)