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21 tháng 1 2019

Ta có:

\(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\dfrac{99+1}{1\cdot99}+\dfrac{97+3}{3\cdot97}+...+\dfrac{1+99}{99\cdot1}}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{\left(1+\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{3}+...+\dfrac{1}{99}+1\right)}{100}}\)

\(=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{100}}=\dfrac{1}{\dfrac{2}{100}}=\dfrac{100}{2}=50\)

21 tháng 1 2019

\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{1+\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{\dfrac{100}{100}+\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)

NV
6 tháng 8 2021

Nếu \(a+b+c=0\)

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+0.\dfrac{2}{abc}\)

\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\)

\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

NV
6 tháng 8 2021

a. Đề bài sai, chắc chắn thiếu dữ kiện \(a+b+c=0\)

b.

\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{\left(-100\right)^2}}=\sqrt{\left(1+\dfrac{1}{99}-\dfrac{1}{100}\right)^2}\)

\(=1+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{9901}{9900}\)

2 tháng 12 2017

Viết gọn lại, ta cần chứng minh:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\right)\)

\(\Leftrightarrow\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{a+b}{ab}}\right)=\sum\dfrac{4ab}{a+b}\)

Thật vậy, ta có:

\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum\left(2\sqrt{\left(a+b\right).\dfrac{1}{4}}\right)^2=\sum a+b\)

Vậy ta cần chứng minh:

\(\sum a+b\ge\sum\dfrac{4ab}{a+b}\Leftrightarrow\sum\left(a+b\right)^2\ge\sum4ab\Leftrightarrow\sum\left(a-b\right)^2\ge0\)

Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c


17 tháng 10 2018

Rút gọn biểu thức chứa căn bậc hai

a) Ta có: \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\dfrac{1}{2}\)

b) Ta có: \(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{99}+\sqrt{100}\)

=-1+10=9

4 tháng 10 2018

Bài 1:Với mọi n∈N*,ta có:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Do đó :

A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

Bài 2: 

\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)

\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)

=10