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Đặt \(\hept{\begin{cases}\sqrt[3]{2-x}=a\\\sqrt[3]{x+7}=b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^2+b^2-ab=3\\a^3+b^3=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2-ab=3\\\left(a+b\right)\left(a^2-ab+b^2\right)=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2-ab=3\\a+b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=1\\b=2\end{cases}}\)hoặc \(\hept{\begin{cases}a=2\\b=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-6\end{cases}}\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
1. ĐKXĐ: $x\geq \frac{-3}{5}$
PT $\Leftrightarrow 5x+3=3-\sqrt{2}$
$\Leftrightarrow x=\frac{-\sqrt{2}}{5}$
2. ĐKXĐ: $x\geq \sqrt{7}$
PT $\Leftrightarrow (\sqrt{x}-7)(\sqrt{x}+7)=4$
$\Leftrightarrow x-49=4$
$\Leftrightarrow x=53$ (thỏa mãn)
Đặt \(\left\{{}\begin{matrix}a=\sqrt[3]{x+7}\\b=\sqrt[3]{2-x}\end{matrix}\right.\) \(\Rightarrow a^3+b^3=9\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=9\) (1)
Pt đã cho tương đương: \(a^2+b^2-ab=3\) (2)
Thay (2) vào (1) ta được:
\(3\left(a+b\right)=9\Rightarrow a+b=3\Rightarrow b=3-a\) (3)
Thay (3) vào (2) ta được:
\(a^2+\left(3-a\right)^2-a\left(3-a\right)-3=0\)
\(\Leftrightarrow a^2+a^2-6a+9-3a+a^2-3=0\) \(\Leftrightarrow3a^2-9a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x+7}=1\\\sqrt[3]{x+7}=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+7=1\\x+7=8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)