d/ \(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{\left(x-1\right)^2}+\sqrt[3]{x^2-1}=1\)
Đặt \(\hept{\begin{cases}\sqrt[3]{x+1}=a\\\sqrt[3]{x-1}=b\end{cases}\Rightarrow a^3-b^3=2}\)
\(\Rightarrow\hept{\begin{cases}a^3-b^3=2\\a^2+b^2+ab=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a^2+b^2+ab\right)=2\\a^2+b^2+ab=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a-b=2\\a^2+b^2+ab=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a-b=2\\b^2+2b+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=1\\b=-1\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=1\\\sqrt[3]{x-1}=-1\end{cases}\Leftrightarrow}x=0}\)
bài b , lập phương lên
bài c , đặt cái căn đưa về hệ
mới nhìn dc làm dc liền thế thui
a/ \(\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1\)
Điều kiện: \(\hept{\begin{cases}x\ge0\\1-x\ge0\end{cases}\Leftrightarrow0\le x\le1}\)
Đặt \(\hept{\begin{cases}\sqrt{x}=a\left(a\ge0\right)\\\sqrt{1-x}=b\left(b\ge0\right)\end{cases}\Rightarrow a^2+b^2=1\left(1\right)}\)
\(PT\Leftrightarrow a+\sqrt{a^2+b}=1\)
\(\Leftrightarrow\sqrt{a^2+b}=1-a\left(a\le1\right)\)
\(\Leftrightarrow a^2+b=a^2-2a+1\)
\(\Leftrightarrow b=1-2a\left(a\le\frac{1}{2}\right)\)
Thế vào (1) ta được
\(a^2+\left(1-2a\right)^2=1\)
\(\Leftrightarrow a^2+4a^2-4a+1=1\)
\(\Leftrightarrow5a^2-4a=0\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{4}{5}\left(l\right)\end{cases}}\)
\(\Rightarrow b=1\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}=0\\\sqrt{1-x}=1\end{cases}\Leftrightarrow x=0}\)
b/ \(\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}\)
\(\Leftrightarrow2x+3\sqrt[3]{x^2-1}\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)=5x\)
\(\Leftrightarrow\sqrt[3]{x^2-1}\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)-x=0\)
Đặt \(\hept{\begin{cases}\sqrt[3]{x+1}=a\\\sqrt[3]{x-1}=b\end{cases}\Rightarrow a^3+b^3=2x}\)
\(\Rightarrow PT\Leftrightarrow ab\left(a+b\right)-\frac{a^3+b^3}{2}=0\)
\(\Leftrightarrow2ab\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2+ab+b^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b=0\\\left(a^2+ab+b^2\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=-b\\a=b=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt[3]{x+1}=-\sqrt[3]{x-1}\\\sqrt[3]{x+1}=\sqrt[3]{x-1}=0\left(l\right)\end{cases}}\)
\(\Leftrightarrow x=0\)
c/ \(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
Điều kiện: \(x+1\ge0\Leftrightarrow x\ge-1\)
Đặt \(\hept{\begin{cases}\sqrt[3]{x-2}=a\\\sqrt{x+1}=b\left(b\ge0\right)\end{cases}\Rightarrow b^2-a^3=3\left(1\right)}\)
\(\Rightarrow PT\Leftrightarrow a+b=3\Leftrightarrow b=3-a\)
Thế vào (1) ta được
\(9-6a+a^2-a^3=3\)
\(\Leftrightarrow a^3-a^2+6a-6=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+6\right)=0\)
\(\Leftrightarrow a=1\Rightarrow b=2\)
\(\Rightarrow\hept{\begin{cases}\sqrt[3]{x-2}=1\\\sqrt{x+1}=2\end{cases}\Leftrightarrow x=3}\)
