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ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !
câu 1 ) thì đúng
câu 2 sai đề
![](https://rs.olm.vn/images/avt/0.png?1311)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(x-2\right)\left(\sqrt[3]{4x-4}+\sqrt{2x-2}\right)=3x-1\)
\(\Leftrightarrow2\left(x-2\right)\left[\left(\sqrt[3]{4x-4}-2\right)+\left(\sqrt{2x-2}-2\right)\right]+8\left(x-2\right)=3x-1\)
\(\Leftrightarrow2\left(x-2\right)\left[\frac{4x-12}{\sqrt[3]{\left(4x-4\right)^2}+2\sqrt[3]{4x-4}+4}+\frac{2x-6}{\sqrt{2x-2}+2}\right]+\left(5x-15=0\right)\)
\(\left(x-3\right)\left[\frac{8\left(x-2\right)}{...}+\frac{4\left(x-2\right)}{...}+5\right]=0\Leftrightarrow x=3.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
toán lớp 9 thì ai mà biết chỉ lớp 5 thôi
đáp án là : 0 bít !
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(3-x\right)\sqrt{x-1}+\sqrt{5-2x}=\sqrt{\left[\left(x-3\right)^2+1\right]\left(4-x\right)}\)
đặt 3-x=a;\(\sqrt{x-1}=b;\sqrt{5-2x}=c\Rightarrow b^2+c^2=4-x\)
\(\Leftrightarrow ab+c=\sqrt{\left(a^2+1\right)\left(b^2+c^2\right)}\)
<=>a2b2+2abc+c2=a2b2+b2+a2c2+c2
<=>b2-2abc+a2c2=0
<=>(b-ac)2=0
<=>b=ac
đến đây thì dễ rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1\)
\(pt\Leftrightarrow\sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}\)
\(\Leftrightarrow\sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)\)
\(\Leftrightarrow\frac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}=\frac{2x^3-3x^2-3x+2}{4x-1}\)
\(\Leftrightarrow\frac{x^3-x^2-2x}{\sqrt{x^3+1}+x+1}-\frac{2x^3-3x^2-3x+2}{4x-1}=0\)
\(\Leftrightarrow\frac{x\left(x-2\right)\left(x+1\right)}{\sqrt{x^3+1}+x+1}-\frac{\left(x+1\right)\left(x-2\right)\left(2x-1\right)}{4x-1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}\right)=0\)
Suy ra x=2;x=-1 còn 1 nghiệm nữa xấu quá t gg :v