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TT
2 tháng 4 2017
Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:
a) Ta có
\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)
\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)
\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)
\(=\sqrt{2017}-1\)
\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)
b) Ta có
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2-1}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Tương tự ta có
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......................
\(\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
Suy ra
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
TT
1 tháng 4 2017
a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5 + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013} + \sqrt {2017} }}\\
= \frac{{(\sqrt 5 + 1)(\sqrt 5 - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017} + \sqrt {2013} )(\sqrt {2017} - \sqrt {2013} )}}{{\sqrt {2013} + \sqrt {2017} }}\\
= \sqrt 5 - 1 + \sqrt 9 - \sqrt 5 + ... + \sqrt {2017} - \sqrt {2013} \\
= 1 + \sqrt 5 - \sqrt 5 + \sqrt 9 - \sqrt 9 + ... + \sqrt {2013} - \sqrt {2013} + \sqrt {2017} \\
= 1 + \sqrt {2017} \\
\Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]
MT
17 tháng 8 2015
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{9}+\sqrt{10}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{10}-\sqrt{9}\)
\(=\sqrt{10}-1\)
\(B=\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{7}}+\frac{2}{\sqrt{7}+\sqrt{9}}\)
\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}\)
\(=\sqrt{9}-1\)
11 tháng 8 2017
Bài 1:
Ta có:
\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)
\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)
\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)
\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)
26 tháng 8 2020
a) Ta có: \(B=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{1-\sqrt{a}}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}}{\sqrt{a}+1}+\frac{\sqrt{a}}{1-a}\right)\)
\(=\left(\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{a+2\sqrt{a}+1-\left(a-2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{a+2\sqrt{a}+1+a-\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{2a+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}{2a+1}\)
\(=\frac{4\sqrt{a}}{2a+1}\)
b) ĐKXĐ: \(0\le a\ne1\)
Ta có: \(a=9-4\sqrt{5}\)
\(=5-2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}-2\right)^2\)(nhận)
Thay \(a=\left(\sqrt{5}-2\right)^2\) vào biểu thức \(B=\frac{4\sqrt{a}}{2a+1}\),ta được:
\(B=\frac{4\cdot\sqrt{\left(\sqrt{5}-2\right)^2}}{2\cdot\left(\sqrt{5}-2\right)^2+1}\)
\(=\frac{4\cdot\left|\sqrt{5}-2\right|}{2\cdot\left(9-4\sqrt{5}\right)+1}\)
\(=\frac{4\cdot\left(\sqrt{5}-2\right)}{18-8\sqrt{5}+1}\)(Vì \(\sqrt{5}>2\))
\(=\frac{4\sqrt{5}-8}{19-8\sqrt{5}}\)
Vậy: Khi \(a=9-4\sqrt{5}\) thì \(B=\frac{4\sqrt{5}-8}{19-8\sqrt{5}}\)
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)