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a) \(x^4-24x+32=0\)
\(\Leftrightarrow x^4-2x^3+2x^3-4x^2+4x^2-8x-16x+32=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)+4x\left(x-2\right)-16\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2+4x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+2x^2+4x-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x\approx1,62\end{matrix}\right.\)
b) \(x^4-8x\sqrt{2}+12=0\)
\(\Leftrightarrow x^4-\sqrt{2}x^3+\sqrt{2}x^3-2x^2+2x^2-2\sqrt{2}x-6\sqrt{2}x+12=0\)
\(\Leftrightarrow x^3\left(x-\sqrt{2}\right)+\sqrt{2}x^2\left(x-\sqrt{2}\right)+2x\left(x-\sqrt{2}\right)-6\sqrt{2}\left(x-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x^3+\sqrt{2}x^2+2x-6\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x\approx1,4142135...\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x\geq -3,5$
PT \(\Leftrightarrow (\sqrt{2x+7}-1)+(\sqrt[3]{x+4}-1)+(x^2+8x+15)=0\)
\(\Leftrightarrow \frac{2(x+3)}{\sqrt{2x+7}+1}+\frac{x+3}{\sqrt[3]{(x+4)^2}+\sqrt[3]{x+4}+1}+(x+3)(x+5)=0\)
\(\Leftrightarrow (x+3)\left[\frac{2}{\sqrt{2x+7}+1}+\frac{1}{\sqrt[3]{(x+4)^2}+\sqrt[3]{x+4}+1}+(x+5)\right]=0\)
Với $x\geq -3,5$ dễ thấy biểu thức trong ngoặc vuông $>0$
Do đó: $x+3=0$
$\Leftrightarrow x=-3$ (thỏa mãn)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
????
xin lỗi nha !
mình mới học lớp 3
mà bài này khó nắm