a) CMR \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\) với \(a\ge0\)và \(a\ne1\).
b) CMR \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}=1\)
K
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31 tháng 7 2020
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
NV
19 tháng 7 2019
cau c í mk thấy bn chép sai đề nên mk sửa lại đề rồi bạn xem lại đề rồi so với bài làm của mk nha có j ko hiểu thì ib mk nha
19 tháng 7 2019
\(a)VT = \dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{a + \sqrt a }}{{\sqrt a }}\\ = \dfrac{{a + 2\sqrt a + 1 - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a }}\\ = \dfrac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)}} + \sqrt a + 1\\ = \dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a - 1}} + \sqrt a + 1\\ = \sqrt a - 1 + \sqrt a + 1\\ = 2\sqrt a = VP (đpcm) \)
\(b)VT = \dfrac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\\ = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - \left( {x - 2\sqrt {xy} + y} \right)\\ = x - \sqrt {xy} + y - x + 2\sqrt {xy} - y\\ = \sqrt {xy} (đpcm)\\ c)VT = \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\ = \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \sqrt a - \sqrt b .\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}\\ = \dfrac{{a - b}}{{a - b}} = 1 (đpcm)\\ d)VT = \left[ {\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2} + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}} \right]:\sqrt b \\ = \dfrac{{a - 2\sqrt {ab} + b + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}:\sqrt b \\ = \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \left( {\sqrt a - \sqrt b } \right):\sqrt b \\ = \sqrt a + \sqrt b - \sqrt a + \sqrt b :\sqrt b \\ = \dfrac{{2\sqrt b }}{{\sqrt b }} = 2 (đpcm) \)
Câu c đề sai (đã sửa)
11 tháng 8 2020
a) Ta có: \(A=\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\left|\sqrt{5}-1\right|}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)(Vì \(\sqrt{5}>1\))
\(=\left(2\sqrt{4+\sqrt{5}-1}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{5+2\cdot\sqrt{5}\cdot1+1}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left|\sqrt{5}+1\right|\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)\)
\(=2\cdot\left(5-1\right)\)
\(=2\cdot4=8\)
b) Ta có: \(B=\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\frac{2}{a+1}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2+\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\right)\cdot\left(\frac{a+1-2}{a+1}\right)^2\)
\(=\frac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2a+2}{\left(a-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2\left(a+1\right)\cdot\left(a-1\right)}{\left(a+1\right)^2}\)
\(=\frac{2a-2}{a+1}\)
LT
8 tháng 8 2019
A= (\(\frac{1}{\sqrt{a}-1}\) - \(\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\)) : \(\frac{\sqrt{a}-2}{a+1}\)
<=> (\(\frac{1}{\sqrt{a-1}}\) - \(\frac{2\sqrt{a}}{\left(a\sqrt{a-1}\right)-\sqrt{a}\left(\sqrt{a}-1\right)}\)). \(\frac{a+1}{\sqrt{a}-2}\)
<=> (\(\frac{1}{\sqrt{a}-1}\) - \(\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1-\sqrt{a}\right)}\)). \(\frac{a+1}{\sqrt{a}-2}\)
<=> (\(\frac{a+1}{\left(\sqrt{a}-1\right)\left(a+1\right)}\).\(\frac{a+1}{\sqrt{a}+2}\)
<=> \(\frac{a+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
a) Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)\)
\(=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\cdot\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\frac{1}{1+\sqrt{a}}\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{1+\sqrt{a}}\)
\(=1+\sqrt{a}\) Bằng 1 kiểu gì đây._.?
a) Xin lỗi sửa lại phần a:
Ta có: \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}\right)^2\cdot\frac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=1\)
b) Ta có: \(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{6}+2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)=3-2=1\)