Question

[Thảo]

giải pt bằng phương pháp liên hợp:

\(\sqrt[3]{x+24}+\sqrt{12-x}=6\)

Answer 1

[Nguyễn Hoàng Minh]

\(ĐK:x\le12\\ PT\Leftrightarrow\left(\sqrt[3]{x+24}-3\right)+\left(\sqrt{12-x}-3\right)=0\\ \Leftrightarrow\dfrac{x-3}{\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9}-\dfrac{x-3}{\sqrt{12-x}+3}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\\dfrac{1}{\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9}=\dfrac{1}{\sqrt{12-x}+3}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt[3]{\left(x+24\right)^2}+3\sqrt[3]{x+24}+9=\sqrt{12-x}+3\\ \Leftrightarrow\sqrt[3]{x+24}\left(\sqrt[3]{x+24}+3\right)+6-\sqrt{12-x}=0\\ \Leftrightarrow\dfrac{\left(x+24\right)\left(\sqrt[3]{x+24}+3\right)}{\sqrt[3]{\left(x+24\right)^2}}+\dfrac{x+24}{6+\sqrt{12-x}}=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-24\left(tm\right)\\\dfrac{\sqrt[3]{x+24}+3}{\sqrt[3]{\left(x+24\right)^2}}=\dfrac{-1}{6+\sqrt{12-x}}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\dfrac{\sqrt[3]{x+24}+3}{\sqrt[3]{x+24}}+\dfrac{1}{\sqrt[3]{x+24}}+\dfrac{1}{6+\sqrt{12-x}}-\dfrac{1}{\sqrt[3]{x+24}}=0\\ \Leftrightarrow\dfrac{\sqrt[3]{x+24}+4}{\sqrt[3]{x+24}}+\dfrac{\sqrt[3]{x+24}+4-10-\sqrt{12-x}}{\sqrt[3]{x+24}\left(6+\sqrt{12-x}\right)}=0\\ \Leftrightarrow\dfrac{x+88}{\sqrt[3]{x+24}\left(\sqrt[3]{\left(x+24\right)^2}-4\sqrt[3]{x+24}+16\right)}+\dfrac{\sqrt[3]{x+24}+4-10-\sqrt{12-x}}{\sqrt[3]{x+24}\left(6+\sqrt{12-x}\right)}=0\)

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 ;-;