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Question

Giải phương trình

1, $\sqrt[3]{12-x}+\sqrt[3]{14+x}=2$

2, $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$

3, $x+\sqrt{17-x^2}+x\sqrt{17-x^2}=9$

4, $\frac{1}{x}+\frac{1}{\sqrt{2-x^2}}=2$

Nguyễn Thành Trương · Accepted Answer

$
1)\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}} = 2\
 \Leftrightarrow 12 - x + 3\sqrt[3]{{{{\left( {12 - x} \right)}^2}.\left( {14 + x} \right)}} + 3\sqrt[3]{{\left( {12 - x} \right){{\left( {14 + x} \right)}^2}}} + 14 + x = 8\
 \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}\left( {\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}}} \right) =  - 18\
 \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}.2 =  - 18\
 \Leftrightarrow \sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}} =  - 3\
 \Leftrightarrow \left( {12 - x} \right)\left( {14 + x} \right) = {\left( { - 3} \right)^3}\
 \Leftrightarrow 168 - 2x - {x^2} =  - 27\
 \Leftrightarrow {x^2} + 2x - 195 = 0\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 15\
x = 13
\end{array} \right.
$

Vậy...Câu trả lời: $
1)\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}} = 2\
 \Leftrightarrow 12 - x + 3\sqrt[3]{{{{\left( {12 - x} \right)}^2}.\left( {14 + x} \right)}} + 3\sqrt[3]{{\left( {12 - x} \right){{\left( {14 + x} \right)}^2}}} + 14 + x = 8\
 \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}\left( {\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}}} \right) =  - 18\
 \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}.2 =  - 18\
 \Leftrightarrow \sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}} =  - 3\
 \Leftrightarrow \left( {12 - x} \right)\left( {14 + x} \right) = {\left( { - 3} \right)^3}\
 \Leftrightarrow 168 - 2x - {x^2} =  - 27\
 \Leftrightarrow {x^2} + 2x - 195 = 0\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 15\
x = 13
\end{array} \right.
$

Vậy...

Nguyễn Thị Ngọc Thơ · Answer

1.

Đặt$\left\{{}\begin{matrix}u=\sqrt[3]{12-x}\v=\sqrt[3]{14+x}\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}u^3=12-x\v^3=14+x\end{matrix}\right.$

$\Rightarrow\left\{{}\begin{matrix}u^3+v^3=26\u+v=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(u+v\right)\left(u^2-uv+v^2\right)=26\u+v=2\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}u^2-uv+v^2=13\v=2-u\end{matrix}\right.$

$\Rightarrow u^2-u\left(2-u\right)+\left(2-u\right)^2=13$ $\Leftrightarrow3u^2-6u-9=0$ $\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=-1\u=-1\Rightarrow v=3\end{matrix}\right.$ Tìm x.

2.ĐK: $-40\le x\le57$

Đặt $\left\{{}\begin{matrix}\sqrt[4]{57-x}=u\\sqrt[4]{x+40}=v\end{matrix}\right.$ $\left(u,v\ge0\right)$ $\Rightarrow\left\{{}\begin{matrix}u^4=57-x\v^4=x+40\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}u+v=5\u^4+v^4=97\end{matrix}\right.$

$\Rightarrow\left\{{}\begin{matrix}u^2+v^2=25-2uv\\left(u^2+v^2\right)^2-2u^2v^2=97\end{matrix}\right.$ $\Rightarrow\left(25-2uv\right)^2-2u^2v^2=97$

$\Leftrightarrow2u^2v^2-100uv+528=0$ $\Rightarrow\left[{}\begin{matrix}uv=44\uv=6\end{matrix}\right.$ Kết hợp $u+v=5$ giải 2 trường hợp.

3.

ĐK: $-\sqrt{17}\le x\le\sqrt{17}$

Đặt $x+\sqrt{17-x^2}=t$ $\Rightarrow\frac{t^2-17}{2}=x\sqrt{17-x^2}$

$PT\Leftrightarrow t+\frac{t^2-17}{2}=9$ $\Leftrightarrow t^2+2t-35=0\Rightarrow\left[{}\begin{matrix}t=5\t=-7\end{matrix}\right.$ Giải tiếp.Câu trả lời: 1.