Mọi người ơi cho hỏi cái phương trình này rút gọn thế nào :
-√x - 2 + x
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\sqrt{x^2y^3}+y\sqrt{x^4y}-xy\sqrt{y}\)
\(=xy\sqrt{y}+x^2y\sqrt{y}-xy\sqrt{y}\)
\(=x^2y\sqrt{y}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\sqrt{\frac{5.\left(38^2-17^2\right)}{8.\left(47^2-19^2\right)}}\)
\(=\sqrt{\frac{5.\left(38-17\right)\left(38+17\right)}{8.\left(47-19\right)\left(47+19\right)}}\)
\(=\sqrt{\frac{5.21.55}{8.28.66}}\)
\(=\sqrt{\frac{5775}{14784}}=\frac{5\sqrt{231}}{2\sqrt{4370}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện xác định : \(x\ge0;y\ge0;\sqrt{x}-\sqrt{y}\ne-3\)
Ta có : \(\frac{x-y+3\sqrt{x}+3\sqrt{y}}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+3\right)}{\sqrt{x}-\sqrt{y}+3}=\sqrt{x}+\sqrt{y}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,M=x^2-64+x^3+64=x^3+x^2\\ b,x=-4\Leftrightarrow M=-64+16=-48\\ c,M=0\Leftrightarrow x^2\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Để pt có 2 nghiê pb thì:
$\Delta'=1-(m-3)>0\Leftrightarrow m< 4$
Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2\\ x_1x_2=m-3\end{matrix}\right.\)
Khi đó:
\(x_1^2-2x_2+x_1x_2=-12\)
\(\Leftrightarrow x_1^2-2(2-x_1)+x_1(2-x_1)=-12\)
\(\Leftrightarrow x_1=-2\Leftrightarrow x_2=2-x_1=4\)
$m-3=x_1x_2=(-2).4=-8$
$\Leftrightarrow m=-5$ (tm)
\(x-\sqrt{x}-2=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)