\(\frac{3}{4x6}\)+\(\frac{3}{6x8}\)+.............+\(\frac{3}{152x154}\)
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\(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).y=\frac{1}{3}\)
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}:\frac{1}{2}=\frac{2}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{2}{3}\)
\(\frac{2}{5}.y=\frac{2}{3}\)
=> \(y=\frac{2}{3}:\frac{2}{5}\)
=>\(y=\frac{3}{5}\)
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Vậy: \(A=\frac{49}{100}\)
Ta có:\(2A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{98.100}\right)\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)
Vậy giá trị của A là \(\frac{49}{200}\)
Ta gọi biểu thức đó là 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
\(\frac{4}{2.4}=2.\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(\frac{4}{4.6}=2.\left(\frac{1}{4}-\frac{1}{6}\right)\)
\(....................\)
\(\frac{4}{18.20}=2.\left(\frac{1}{18}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{18}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(\Rightarrow\)\(A=2.\left(\frac{9}{20}\right)=\frac{18}{20}\)
Ai thấy đúng thì ủng hộ nha !!!
ax4=1/2-1/4+1/4-1/6+1/6+1/8+.......+1/16-1/18+1/18+1/20
ax4 =1/2-1/20
ax4 =9/20
a=9/20:4
a=9/80
Đặt : \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{10}-\frac{1}{12}\)
\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow A=\frac{5}{12}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
B = \(\frac{3^2}{2.4}+\frac{3^2}{4.6}+\frac{3^2}{6.8}+...+\frac{3^2}{198.200}\)
B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{3^2}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{3^2}{2}.\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{3^2}{2}.\left(\frac{1}{198}-\frac{1}{200}\right)\)
B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{198}-\frac{1}{200}\right)\)
B = \(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{200}\right)\)
B = \(\frac{9}{2}.\frac{99}{200}\)
B = \(\frac{891}{400}\)
D = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 48 x 49
3D = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + 4 x 5 x 3 + ... + 48 x 49 x 3
3D = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + 4 x 5 x ( 6 - 3 ) + ... + 48 x 49 x ( 50 - 47 )
3D = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 48 x 49 x 50 - 47 x 48 x 49
3D = 48 x 49 x 50
D = ( 48 x 49 x 50 ) : 3
D = 39200
E = 12 + 22 + 32 + ... + 482
E = 1 x 1 + 2 x 2 + 3 x 3 + ... + 48 x 48
E = 1 x ( 2 - 1 ) + 2 x ( 3 - 1 ) + 3 x ( 4 - 1 ) + ... + 48 x ( 49 - 1 )
E = 1 x 2 - 1 + 2 x 3 - 2 + 3 x 4 - 3 + ... + 48 x 49 - 49
E = ( 1 x 2 + 2 x 3 + 3 x 4 + ... + 48 x 49 ) - ( 1 + 2 + 3 + ... + 49 )
Ta tính được vế trong ngoặc thứ nhất là 39200 , còn vế trong ngoặc thứ hai là 1225
thay vào ta được :
E = 39200 - 1225
E = 37975
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{100}}\)
Đặt A =\(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+...+\frac{1}{2014x2016}\)
A x 2 = \(\frac{2}{2x4}+\frac{2}{4x6}+\frac{2}{6x8}+...+\frac{2}{2014x2016}\)
A x 2 = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\)
A x 2 = \(\frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)
A = \(\frac{1007}{2016}:2=\frac{1007}{4032}\)
Đáp số: \(\frac{1007}{4032}\)
Gọi tông trên là A
2A=2/2.4+2/4.6+2/6.8+........+2/2014.2016
2A=1/2-1/4+1/4-1/6+.........+1/2014-1/2016
2A=1/2-(1/4-1/4)+(1/6-1/6)+...........+(1/2014-1/2014)-1/2016
2A=1/2-1/2016
2A=1007/2016
A=1007/4032
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)
=1/2-1/4+1/4-1/6+1/6-1/8+...+1/2014-1/2016
=1/2-1/2016
=1007/2016
TK va ket ban nhe