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\(\sqrt[3]{9-x}+\sqrt{x+3}=4\)
\(ĐK:x\ge-3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{9-x}=t\\\sqrt{x+3}=p\end{matrix}\right.\); \(p\ge0\) \(\Rightarrow\left\{{}\begin{matrix}t^3=9-x\\p^2=x+3\end{matrix}\right.\) \(\Leftrightarrow t^3+p^2=12\)
Ta có hpt: \(\left\{{}\begin{matrix}t+p=4\\t^3+p^2=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}p=4-t\\t^3+p^2=12\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow t^3+\left(4-t\right)^2=12\)
\(\Leftrightarrow t^3+t^2-8t+16=12\)
\(\Leftrightarrow t^3+t^2-8t+4=0\)
\(\Leftrightarrow t^3-2t^2+3t^2-8t+4=0\)
\(\Leftrightarrow t^2\left(t-2\right)+\left(3t^2-6t-2t+4\right)=0\)
\(\Leftrightarrow t^2\left(t-2\right)+3t\left(t-2\right)-2\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t^2+3t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t^2+3t-2=0\left(3\right)\end{matrix}\right.\)
\(\left(3\right)t^2+3t-2=0\)
\(\Delta=3^2-4.\left(-2\right)=17>0\)
\(\rightarrow\left[{}\begin{matrix}t=\dfrac{-3+\sqrt{17}}{2}\\t=\dfrac{-3-\sqrt{17}}{2}\end{matrix}\right.\)
`@` TH1: \(t=2\Rightarrow p=4-2=2\left(tm\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{9-x}=2\\\sqrt{x+3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9-x=8\\x+3=4\end{matrix}\right.\) \(\Leftrightarrow x=1\left(tm\right)\)
`@`TH2: \(t=\dfrac{-3\pm\sqrt{17}}{2}\) \(\Rightarrow p=4-\dfrac{-3\pm\sqrt{17}}{2}=\dfrac{11\pm\sqrt{17}}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{9-x}=\dfrac{-3\pm\sqrt{17}}{2}\\\sqrt{x+3}=\dfrac{11\pm\sqrt{17}}{2}\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{63\pm11\sqrt{17}}{2}\left(tm\right)\)
Vậy \(S=\left\{1;\dfrac{63\pm11\sqrt{17}}{2}\right\}\)
\(\dfrac{x}{x+1}-2\sqrt{\dfrac{x+1}{x}}=3\)
\(ĐK:\sqrt{\dfrac{x+1}{x}}\ge0\)
Đặt \(\sqrt{\dfrac{x+1}{x}}=t;t\ge0\)
\(\Leftrightarrow\dfrac{x+1}{x}=t^2\) \(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{1}{t^2}\)
Ptr trở thành:
\(\Leftrightarrow\dfrac{1}{t^2}-2t=3\)
\(\Leftrightarrow\dfrac{1-2t^3}{t^2}=3\)
\(\Leftrightarrow1-2t^3=3t^2\)
\(\Leftrightarrow2t^3+3t^2-1=0\)
\(\Leftrightarrow2t^3+2t^2+t^2-1=0\)
\(\Leftrightarrow2t^2\left(t+1\right)+\left(t-1\right)\left(t+1\right)=0\)
\(\Leftrightarrow\left(t+1\right)\left(2t^2+t-1\right)=0\)
\(\Leftrightarrow2t^2+t-1=0\) ( vì \(t+1>0\) )
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\left(ktm\right)\\t=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\) ( Vi-ét )
`<=>`\(\sqrt{\dfrac{x+1}{x}}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x}=\dfrac{1}{4}\)
\(\Leftrightarrow4x+4=x\)
\(\Leftrightarrow3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)
ĐKXĐ: \(x\ge-3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{9-x}=a\\\sqrt{x+3}=b\ge0\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+b=4\\a^3+b^2=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=4-b\\a^3+b^2=12\end{matrix}\right.\)
\(\Rightarrow\left(4-b\right)^3+b^2=12\)
\(\Leftrightarrow b^3-13b^2+48b-52=0\)
\(\Leftrightarrow\left(b-2\right)\left(b^2-11b+26\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=2\\b=\dfrac{11-\sqrt{17}}{2}\\b=\dfrac{11+\sqrt{17}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{x+3}=\dfrac{11-\sqrt{17}}{2}\\\sqrt{x+3}=\dfrac{11+\sqrt{17}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{63-11\sqrt{17}}{2}\\x=\dfrac{63+11\sqrt{17}}{2}\end{matrix}\right.\)
