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\(ĐK:x\le12\\ PT\Leftrightarrow\left(\sqrt[3]{x+24}-3\right)+\left(\sqrt{12-x}-3\right)=0\\ \Leftrightarrow\dfrac{x-3}{\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9}-\dfrac{x-3}{\sqrt{12-x}+3}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\\dfrac{1}{\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9}=\dfrac{1}{\sqrt{12-x}+3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9=\sqrt{12-x}+3\\ \Leftrightarrow\sqrt[3]{x+24}\left(\sqrt[3]{x+24}+3\right)+6-\sqrt{12-x}=0\\ \Leftrightarrow\dfrac{\left(x+24\right)\left(\sqrt[3]{x+24}+3\right)}{\sqrt[3]{\left(x+24\right)^2}}+\dfrac{x+24}{6+\sqrt{12-x}}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-24\left(tm\right)\\\dfrac{\sqrt[3]{x+24}+3}{\sqrt[3]{\left(x+24\right)^2}}=\dfrac{-1}{6+\sqrt{12-x}}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\dfrac{\sqrt[3]{x+24}+3}{\sqrt[3]{x+24}}+\dfrac{1}{\sqrt[3]{x+24}}+\dfrac{1}{6+\sqrt{12-x}}-\dfrac{1}{\sqrt[3]{x+24}}=0\\ \Leftrightarrow\dfrac{\sqrt[3]{x+24}+4}{\sqrt[3]{x+24}}+\dfrac{\sqrt[3]{x+24}+4-10-\sqrt{12-x}}{\sqrt[3]{x+24}\left(6+\sqrt{12-x}\right)}=0\\ \Leftrightarrow\dfrac{x+88}{\sqrt[3]{x+24}\left(\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16\right)}+\dfrac{\sqrt[3]{x+24}+4-10-\sqrt{12-x}}{\sqrt[3]{x+24}\left(6+\sqrt{12-x}\right)}=0\)
Xét \(\sqrt[3]{x+24}+4-10-\sqrt{12-x}=\dfrac{x+88}{\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16}-\dfrac{x+88}{10+\sqrt{12-x}}=0\)
\(=\left(x+88\right)\left(\dfrac{1}{\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16}-\dfrac{1}{10+\sqrt{12-x}}\right)\)
Thay vào PT (2) ta đặt đc nhân tử chung là \(x+88\)
Và ngoặc lớn còn lại vô nghiệm
\(\Leftrightarrow x+88=0\Leftrightarrow x=-88\left(tm\right)\)
Vậy PT có nghiệm \(x\in\left\{-88;-24;3\right\}\)
P/s mình thấy giải theo PP đặt ẩn phụ dễ hơn á ;-;
a/ ĐKXĐ: ...
\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)
\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))
\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-3\)
\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)
\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\le12\)
\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)
\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)
\(\Leftrightarrow a^3+a^2-12a=0\)
\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)
\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
đặt \(x^2+5x+5=t\)
\(\Leftrightarrow t^2-25=0\Rightarrow\left\{{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
