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Đặt A
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{5}{2}.\left(\frac{12}{25}\right)=\frac{6}{5}\)
Ai thấy đúng thì ủng hộ nha !!!
a, \(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
=\(\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)=\(\frac{5}{2}.\frac{12}{25}\)=\(\frac{6}{5}\)
\(F=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(F=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(F=2\times\frac{502}{1005}\)
\(F=\frac{1004}{1005}\)
F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010
F=2/2-2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2- 2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2-2/2010
=>F=2008/2010=1004/1005
\(\left(\frac{4}{2.4}+\frac{4}{4.6}+.....+\frac{4}{4020.4022}\right)x=2010\)
\(\Leftrightarrow2x\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{4020.4022}\right)=2010\)
\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4}+.....+\frac{1}{4020}-\frac{1}{4022}\right)=2010\)
\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4022}\right)=2010\)
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
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
Đặt \(A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
Ủng hộ mk nha !!! *_*
\(\text{Đ}\text{ặt}:A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
vậy 4A=4/2-4/4-4/6+...+4/46-4/48-4/50
4A=48/25
A=12/25