Cho biểu thức:Q=\(\left(\frac{1}{\sqrt{x}}+\frac{5\sqrt{x}-4}{2\sqrt{x}-x}\right):\left(\frac{2+\sqrt{x}}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}-2}\right)\)
a,rút gọn Q
b,Tìm m để x thỏa mãn Q=\(mx\sqrt{x}-2mx+1\)
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VH
22 tháng 8 2020
P/s : sửa đề
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}-3x}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{-3\sqrt{x}}{\sqrt{x}+3}\)
b) \(P< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6\sqrt{x}+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-5\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
Mà \(2\left(\sqrt{x}+3\right)>0\)
\(\Rightarrow-5\sqrt{x}+3< 0\)
\(\Leftrightarrow-5\sqrt{x}< -3\)
\(\Leftrightarrow\sqrt{x}>\frac{3}{5}\)
\(\Leftrightarrow x>\frac{9}{25}\)
Vấy .................
VH
22 tháng 8 2020
c) \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}-2-2+x=0\)
\(\Leftrightarrow-\sqrt{x}-4+x=0\)
\(\Leftrightarrow-\sqrt{x}\left(1-\sqrt{x}\right)=4\)
Còn lại lập bảng tự tìm giá trị của x là ra .( Chú ý : đối chiếu ĐKXĐ )
d)
\(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+x\sqrt{x}-xm=x-3\sqrt{x}-m\sqrt{x}\)
\(\Leftrightarrow-3\sqrt{x}+x\sqrt{x}-xm-x+3\sqrt{x}+m\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x+m\right)-x\left(m+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left[x+m-m\sqrt{x}-\sqrt{x}\right]=0\)
\(\Leftrightarrow\sqrt{x}\left[m\left(1-\sqrt{x}\right)-\sqrt{x}\left(1-\sqrt{x}\right)\right]=0\)
\(\Leftrightarrow\sqrt{x}=0;m-\sqrt{x}=0;1-\sqrt{x}=0\)
+) \(\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
+) \(1-\sqrt{x}=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
+) \(m-\sqrt{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-\sqrt{0}=0\\m-\sqrt{1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\m=1\end{cases}}}\)
Vậy ..................
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
19 tháng 8 2019
\(đkxđ\Leftrightarrow x\ge0\)
\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}}:\frac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(x-1\right)\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}.\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-1}{\sqrt{x}}\)
\(b,P.\sqrt{x}=6\sqrt{x}-3-\sqrt{x}-4\)
\(\Rightarrow\frac{x-1}{\sqrt{x}}.\sqrt{x}=5\sqrt{x}-7\)
\(\Rightarrow x-5\sqrt{x}+6=0\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=3\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=9\end{cases}}}\)
Vậy \(x\in\left\{4;9\right\}\)