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\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(x-7\right)}{\left(x-4\right)\left(x-7\right)}=\frac{\left(x+5\right)\left(x-4\right)}{\left(x-4\right)\left(x-7\right)}\)
\(\Rightarrow\)\(^{x^2-7x+x-7=x^2-4x+5x-20}\)
\(\Leftrightarrow\)\(x^2-x^2-6x-x-7+20=0\)
\(\Leftrightarrow-7x+13=0\)
\(\Leftrightarrow-7x=-13\)
\(\Rightarrow x=\frac{13}{7}\)
\(pt\Leftrightarrow\left(x+1\right)\left(x-7\right)=\left(x-4\right)\left(x+5\right)\)
\(\Leftrightarrow x^2-7x+x-7=x^2+5x-4x-20\)
\(\Leftrightarrow-7x=-13\Rightarrow x=\frac{13}{7}\)
\(\frac{1}{11^2}+\frac{1}{12^2}+\frac{1}{13^2}+\frac{1}{14^2}+...+\frac{1}{100^2}\)
\(=\frac{1}{11.11}+\frac{1}{12.12}+\frac{1}{13.13}+\frac{1}{14.14}+...+\frac{1}{100.100}\)
\(< \frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}\)
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 \(\frac{1}{10}=0,1\)
\(\Rightarrow0,08521515902< 0,1\)
thực ra nó rất là dễ. giờ mình mới phát hiện ra chứ bữa trước mình làm cách dài lắm
Ta có :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)\)
\(=\frac{25}{12}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)>\frac{25}{12}\)( đpcm )
Ta có:
\(\frac{1}{12}>\frac{1}{20}\)
\(\frac{1}{13}>\frac{1}{20}\)
\(\frac{1}{14}>\frac{1}{20}\)
......
\(\frac{1}{19}>\frac{1}{20}\)
\(\Rightarrow\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}\)\(>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)
\(=\frac{8}{20}=\frac{2}{5}>\frac{1}{3}\)
\(\Rightarrow\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}>\frac{1}{3}\)
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
\(B=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\\
=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\\
=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\\
=-\left(1-\frac{1}{100}\right)=\frac{-99}{100}\)