\(\sqrt[3]{x-20}+\sqrt{x+15}=7\)
\(ĐK:x\ge-15\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-20}=a\\\sqrt{x+15}=b\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}a^3=x-20\\b^2=x+15\end{matrix}\right.\)
Ta có hpt: \(\left\{{}\begin{matrix}a+b=7\\a^3-b^2=-35\left(2\right)\end{matrix}\right.\) \(\rightarrow b=7-a\)
\(\left(2\right)a^3-b^2=-35\)
\(\Leftrightarrow a^3-\left(7-a\right)^2=-35\)
\(\Leftrightarrow a^3-a^2+14a-49+35=0\)
\(\Leftrightarrow a^3-a^2+14a-14=0\)
\(\Leftrightarrow a^2\left(a-1\right)+14\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+14\right)=0\)
\(\Leftrightarrow a-1=0\)
\(\Leftrightarrow a=1\)
\(\rightarrow b=7-1=6\)
\(\rightarrow\left\{{}\begin{matrix}\sqrt[3]{x-20}=1\\\sqrt{x+15}=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-20=1\\x+15=36\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=21\\x=21\end{matrix}\right.\) ( tm )
Vậy \(S=\left\{21\right\}\)
ĐK :`x>=-15`
\(\left\{{}\begin{matrix}\sqrt{x+15}=a>0\\\sqrt[3]{x-20}=b\end{matrix}\right.\) ta có hệ :
\(\left\{{}\begin{matrix}a+b=7\\a^2-b^3=35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7-b\\a^2-b^3=35\end{matrix}\right.\\ \Rightarrow\left(7-b\right)^2-b^3=35\\ \Rightarrow-b^3+b^2-14b+14=0\\ \Leftrightarrow b^2\left(1-b\right)+14\left(1-b\right)=0\\ \Leftrightarrow\left(b^2+14\right)\left(1-b\right)=0\\ \Rightarrow b=1\\ \Rightarrow\sqrt[3]{x-20}=1\\ \Rightarrow x=21\)
