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Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
\(\left(\frac{4}{2.4}+\frac{4}{4.6}+.....+\frac{4}{4020.4022}\right)x=2010\)
\(\Leftrightarrow2x\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{4020.4022}\right)=2010\)
\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4}+.....+\frac{1}{4020}-\frac{1}{4022}\right)=2010\)
\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4022}\right)=2010\)
Tự biên tự diễn
Ko chép lại đề nhé
<=> 2( 2/2.4 + 2/2.6 + 2/2.8 +...+ 2/ 4020.4022) x= 2010
<=> 2( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +....+ 1/4020- 1/4022 )x=2010
<=> ( 1/2 - 1/4022)2x = 2010
<=> ( 2011/4022 - 1/4022 )2x = 2010
<=>( 2010/4022) .2x= 2010
<=> 2x = 2010 : 2010/4022
<=> 2x = 4022
=> x = 2011
Vậy x = 2011
đụ cha mi
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mấy bài này cũng dễ ẹt nữa
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\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)
ĐKXĐ: \(x\ne0;x\ne-2\)
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+2}=\frac{1}{18}\)
\(\Rightarrow\)\(x+2=18\)
\(\Leftrightarrow\)\(x=16\) (t/m ĐKXĐ)
Vậy...
1/2(1-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2)=4/9
1/2(1-1/x+2)=4/9
1- 1/x+2=4/9:1/2
1 - 1 /x+2=8/9
1/x+2=1-8/9
1/x+2=1/9
suy ra x+2=9
x=9-2
x=7