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Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
B1:
\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=\sqrt{3}+2\sqrt{2}\)
\(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{4-4\sqrt{3}+3}+\left|1+\sqrt{3}\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}\)
\(=2-\sqrt{3}+1+\sqrt{3}\)
\(=3\)
B2:
đk: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\frac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
\(\Leftrightarrow x+2=9\)
\(\Rightarrow x=7\)
Vậy x = 7
B3:
a) đk: \(\hept{\begin{cases}x>-1\\x\ne\left\{0;1\right\}\end{cases}}\)
b) Ta có:
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(P=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}\div\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(P=\frac{x-\sqrt{x}}{\sqrt{x}+1}\)
2) \(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow x=7\left(x\ge-2\right)\)
3) ĐK: \(x\ge0,x\ne1\)
Ta có: \(P=\left[\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{x-1}\right]\)
\(=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}:\frac{\sqrt{x}-1+2}{x-1}\)\(=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}.\frac{x-1}{\sqrt{x}+1}=\frac{\left(x-1\right)^2}{\left(x-1\right)\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
Bài 1.
1) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{2\cdot9}\)
\(=-\left(\sqrt{2}-\sqrt{3}\right)+\sqrt{2\cdot3^2}\)
\(=\sqrt{3}-\sqrt{2}+\left|3\right|\sqrt{2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
2) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{3-4\sqrt{3}+4}+\left|1+\sqrt{3}\right|\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot2+2^2}+1+\sqrt{3}\)
\(\sqrt{\left(\sqrt{3}-2\right)^2}+1+\sqrt{3}\)
\(=\left|\sqrt{3}-2\right|+1+\sqrt{3}\)
\(=-\left(\sqrt{3}-2\right)+1+\sqrt{3}\)
\(=2-\sqrt{3}+1+\sqrt{3}=3\)
Bài 2.
\(\sqrt{9x+18}-5\sqrt{x+2}+\frac{4}{5}\sqrt{25x+50}=6\)
ĐK : x ≥ -2
<=> \(\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\frac{4}{5}\sqrt{25\left(x+2\right)}=6\)
<=> \(\sqrt{3^2\left(x+2\right)}-5\sqrt{x+2}+\frac{4}{5}\sqrt{5^2\left(x+2\right)}=6\)
<=> \(\left|3\right|\sqrt{x+2}-5\sqrt{x+2}+\frac{4}{5}\cdot\left|5\right|\sqrt{x+2}=6\)
<=> \(3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
<=> \(\sqrt{x+2}\left(3-5+4\right)=6\)
<=> \(\sqrt{x+2}\times2=6\)
<=> \(\sqrt{x+2}=3\)
<=> \(x+2=9\)
<=> \(x=7\)( tm )
Bài 3.
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\sqrt{x}\times\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{1\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\div\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\times\left(\sqrt{x}-1\right)=\frac{x-1}{\sqrt{x}}\)