ĐKXĐ: \(x\ge-1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\ge1\\\sqrt{x+1}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a-b=1\\a^3-b^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a-1\\a^3-b^2=1\end{matrix}\right.\)
\(\Rightarrow a^3-\left(a-1\right)^2=1\)
\(\Leftrightarrow a^3-a^2+2a=0\)
\(\Leftrightarrow a\left(a^2-a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(l\right)\\a^2-a+2=0\left(vn\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
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