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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(x-7\right)}{\left(x-4\right)\left(x-7\right)}=\frac{\left(x+5\right)\left(x-4\right)}{\left(x-4\right)\left(x-7\right)}\)
\(\Rightarrow\)\(^{x^2-7x+x-7=x^2-4x+5x-20}\)
\(\Leftrightarrow\)\(x^2-x^2-6x-x-7+20=0\)
\(\Leftrightarrow-7x+13=0\)
\(\Leftrightarrow-7x=-13\)
\(\Rightarrow x=\frac{13}{7}\)
\(pt\Leftrightarrow\left(x+1\right)\left(x-7\right)=\left(x-4\right)\left(x+5\right)\)
\(\Leftrightarrow x^2-7x+x-7=x^2+5x-4x-20\)
\(\Leftrightarrow-7x=-13\Rightarrow x=\frac{13}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{11^2}+\frac{1}{12^2}+\frac{1}{13^2}+\frac{1}{14^2}+...+\frac{1}{100^2}\)
\(=\frac{1}{11.11}+\frac{1}{12.12}+\frac{1}{13.13}+\frac{1}{14.14}+...+\frac{1}{100.100}\)
\(< \frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}\)
Vì \(\frac{1}{100}>0\Rightarrow\frac{1}{10}-\frac{1}{100}< \frac{1}{10}\)
\(\RightarrowĐPCM\)
theo mình tình thi \(\frac{1}{11^2}+\frac{1}{12^2}+......+\frac{1}{100^2}=0,08521616902\)
mà \(\frac{1}{10}=0,1\)
\(\Rightarrow0,08521515902< 0,1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Rút gọn phân số :
\(-\frac{28}{63}=-\frac{4}{9}\)
\(\frac{-105}{-770}=\frac{3}{22}\)
\(\frac{12+19}{-75-18}=\frac{31}{-93}=-\frac{1}{3}\)
\(BCNN\left(9;22;3\right)=198\)
Quy đồng 3 phân số :
\(-\frac{4}{9}=-\frac{4.\left(198:9\right)}{198}=-\frac{88}{198}\)
\(\frac{3}{22}=\frac{3.\left(198:22\right)}{198}=\frac{27}{198}\)
\(-\frac{1}{3}=\frac{1\left(198:3\right)}{198}\frac{66}{198}\)
\(\frac{-28}{63}\)=\(\frac{-28:7}{63:7}\)=\(\frac{-4}{9}\)
\(\frac{-105}{770}=\frac{-105:35}{770:35}=\frac{-3}{22}\)
\(\frac{12+19}{-75-18}=\frac{31}{-93}=\frac{31:31}{-93:31}=\frac{1}{-3}=-\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
A= \(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{22}>\) \(\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=\frac{11}{22}=\frac{1}{2}\)
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11 số 1/22
Từ trên ta có đpcm