cho biểu thức P=\(\left[\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)
a/ Rút gọn P
b/ Tìm a để \(\frac{1}{P}-\frac{\sqrt{a}+1}{8}\ge1\)
K
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5 tháng 10 2019
a/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}\)
b/ \(\sqrt{ab}+\sqrt{a}-\sqrt{b}=2\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)\left(\sqrt{b}+1\right)=1\)
Xong rồi nhá
NV
14 tháng 7 2016
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
14 tháng 7 2016
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
1 tháng 8 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(A=\)\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}.\sqrt{a}}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)^2\)\(\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{\left(2\sqrt{a}\right)^2}\left(\frac{a-2\sqrt{a}+1-a-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2.-4\sqrt{a}}{4a\left(a-1\right)}=\frac{a-1}{\sqrt{a}}\)
\(b,A< 0\Rightarrow\frac{a-1}{\sqrt{a}}< 0\)
Mà \(\sqrt{a}\ge0\Rightarrow a-1\le0\Rightarrow a\le1\)
\(A=2\Rightarrow\frac{a-1}{\sqrt{a}}=2\)
\(\Rightarrow a-1=2\sqrt{a}\Rightarrow a-2\sqrt{a}-1=0\)
\(\Rightarrow a-2\sqrt{a}+1-2=0\)
\(\Rightarrow\left(\sqrt{a}-1\right)^2-\sqrt{2}^2=0\)
\(\Rightarrow\left(\sqrt{a}-1-\sqrt{2}\right)\left(\sqrt{a}-1+\sqrt{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=1+\sqrt{2}\\\sqrt{a}=1-\sqrt{2}\end{cases}\Rightarrow\orbr{\begin{cases}a=\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}\\a=\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}\end{cases}}}\)
BA
1 tháng 8 2019
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(a-1\right)^2}{4a}.\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{a-1}\)
\(=\frac{a-1}{4a}.\frac{2\sqrt{a}.\left(-2\right)}{1}\)
\(=\frac{a-1}{4a}.\frac{-4\sqrt{a}.}{1}\)
\(=\frac{1-a}{\sqrt{a}}\)
13 tháng 8 2019
\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)
Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)