1, tìm x bt :
a,\(\sqrt{x}\)= 10
b,\(\sqrt{x}\)=\(\frac{1}{3}\)
c,\(\sqrt{x}\)= 9
K
Khách
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TH
13 tháng 10 2015
a/
\(=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)^2}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
b/ Vậy để P>1 khi BT trên>1
Ta có phương trình tương đương
\(x-3\sqrt{x}+3-x\sqrt{x}+6\text{x}-9>0\)
\(-x\sqrt{x}+7\text{x}-3\sqrt{x}-6>0\)
Giải pt rồi suy ra
tick cho mình nha
YH
25 tháng 8 2019
...ghi lại đề...
ĐK: \(x\ge0\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{9-x}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\frac{3\sqrt{x}-x+x+9}{9-x}:\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{3-\sqrt{x}}:\frac{2\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)}=\frac{-3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
Học tốt!
21 tháng 2 2022
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{2\sqrt{x}+x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)
KT
24 tháng 7 2018
a) ĐKXĐ: \(x>0;x\ne9\)
\(A=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+3}\)
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
21 tháng 6 2019
\(a,\sqrt{4-4x+x^2}+\sqrt{\frac{2}{x^2+6x+9}}=\sqrt{\left(x-2\right)^2}+\sqrt{\frac{2}{\left(x+3\right)^2}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x+2\ge0\\x+3>0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-2\\x>-3\end{cases}\Rightarrow}x\ge-2}\)
\(b,\frac{5\sqrt{x}}{\sqrt{x}-3}+\frac{2}{\sqrt{x}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\sqrt{x}-3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\ne\sqrt{9}\end{cases}\Rightarrow}\hept{\begin{cases}x>0\\x\ne9\end{cases}}}\)
\(c,\sqrt{3-\sqrt{x}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\3-\sqrt{x}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\le3\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\le9\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x\le3\end{cases}}}\)
\(\Rightarrow0< x\le3\)
a)x=100
b)x=1/9
c)x=81
a) x = 100
b) x = 1/9
c) x = 81
chúc bạn hok tốt!!!!!!!!