1.
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+24}=b\\\sqrt{12-x}=a\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^3=36\end{matrix}\right.\)
\(\Rightarrow b^3+\left(6-b\right)^2=36\)
\(\Leftrightarrow b^3+b^2-12b=0\)
\(\Leftrightarrow b\left(b^2+b-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=0\\b=3\\b=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+24=0\\x+24=27\\x+24=-64\end{matrix}\right.\)
2.
\(\Leftrightarrow x^2-x+4-\sqrt{x^2-x+4}-2=0\)
Đặt \(\sqrt{x^2-x+4}=t>0\)
\(t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-x+4}=2\)
\(\Leftrightarrow x^2-x=0\)
3.
Bạn coi lại đề
4.
ĐKXĐ: ...
Đặt \(\sqrt{x+3}=a\ge0\)
\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)
\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)
\(\Rightarrow2x^2-ax-a^2=0\)
\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=x\\a=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)
5.
\(\Leftrightarrow x^2+7-\left(x+4\right)\sqrt{x^2+7}+4x=0\)
Đặt \(\sqrt{x^2+7}=t>0\)
\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)
\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+4+x-4}{2}=x\\t=\frac{x+4-x+4}{2}=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vn\right)\\x^2+7=16\end{matrix}\right.\)
Câu 6 bạn coi lại đề
