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12 tháng 7 2016
a) ĐKXĐ : \(0\le a\ne1\)
\(\frac{\sqrt{a}-a}{\sqrt{a}-1}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
b) ĐKXĐ : \(b\ne0,a\ne-\sqrt{b}\)
\(\frac{a-\sqrt{b}}{\sqrt{b}}:\frac{\sqrt{b}}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\sqrt{b}}.\frac{a+\sqrt{b}}{\sqrt{b}}=\frac{a^2-b}{b}=\frac{a^2}{b}-1\)
c) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=\sqrt{5}\left(2-5-4+11\right)\)\(=4\sqrt{5}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
\(=7\left(2-2\sqrt{2}+1\right)+14\sqrt{2}=7\left(2-2\sqrt{2}+1+2\sqrt{2}\right)=7.3=21\)
e) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
1 tháng 8 2023
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
5 tháng 4 2021
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
5 tháng 4 2021
b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
15 tháng 10 2017
a) B= \(\frac{1}{\sqrt{a}}\)(ĐKXĐ: a,b>0) B) Khi a= \(6+2\sqrt{5}\)thì B=\(\frac{1}{\sqrt{\left(\sqrt{5}+1\right)^2}}\)=\(\frac{1}{\sqrt{5}+1}\) C) Do \(\sqrt{a}>0\)\(\Rightarrow\frac{1}{\sqrt{a}}>0\)\(\Rightarrow\frac{1}{\sqrt{a}}>-1\)
30 tháng 8 2018
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{1}{\sqrt{2}}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
1 tháng 9 2016
a) \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(ĐK:a>0;a\ne1;a\ne4\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để \(A>\frac{1}{6}\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Leftrightarrow\)\(\frac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
\(\Leftrightarrow\sqrt{a}-4>0\Leftrightarrow a>16\left(tm\right)\)
Vậy a>16 thì \(A>\frac{1}{6}\)
1 tháng 9 2016
ĐKXĐ : \(a>0,a\ne4,a\ne1\)
a) \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) \(A>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow-\frac{2}{3\sqrt{a}}+\frac{1}{3}>\frac{1}{6}\Rightarrow\frac{2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{1}{\sqrt{a}}>\frac{1}{4}\Rightarrow a< 16\)
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NT
bài 1 rút gon biểu thức
\(6\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+2\sqrt{\dfrac{1}{6}}\)
2
TA
24 tháng 6 2021
`6\sqrt(2/3)-\sqrt(24)+2\sqrt(3/8)+2\sqrt(1/6)`
`=6. \sqrt6/3 - \sqrt(2^2 .6) + 2. \sqrt(24)/8 + 2. \sqrt6/6`
`=2\sqrt6-2\sqrt6+ \sqrt6/2 + \sqrt6/3`
`=\sqrt6/2+\sqrt6/3`
`=(3\sqrt6+2\sqrt6)/6`
`=(5\sqrt6)/6`
24 tháng 6 2021
Ta có: \(6\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+2\sqrt{\dfrac{1}{6}}\)
\(=\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}+2\cdot\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{2}{\sqrt{6}}\)
\(=2\sqrt{6}-2\sqrt{6}+\dfrac{\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{6}}{3}=\dfrac{5\sqrt{6}}{6}\)
\(A=\frac{2\left(\sqrt{7}+\sqrt{6}\right)}{1}-2\sqrt{7}+3\sqrt{6}\)
\(=-\sqrt{6}\)
Học tốt!!!!!!!!!!!!!!!!!!!!!!!