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Bài làm
\(M=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{98.101}\)
\(M=3^2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(M=9\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(M=9.\frac{101-2}{202}\)
\(M=9.\frac{99}{202}\)
\(M=\frac{891}{202}\)
Vậy \(M=\frac{891}{202}\)
\(M=3\left(\frac{3}{2x5}+\frac{3}{5x8}+\frac{3}{8x11}+...+\frac{3}{98.101}\right).\)
\(M=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{101-98}{98.101}\right)\)
\(M=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)=3\left(\frac{1}{2}-\frac{1}{101}\right)=\frac{3.99}{202}\)
A= 1+(2-3)+(5-4)+...+(98-99)-100
=1-1+1-1+...+1-1-100
=-100
Ta có: C=1 - 2 - 3 - 4 + 5 - 6 - 7 - 8 + 9 - 10 - 11 - 12 +... + 97 - 98 - 99 - 100
C= ( 1 - 2 - 3 - 4 ) + ( 5 - 6 - 7 - 8 ) + ( 9 - 10 - 11 - 12 ) + ... + ( 97 - 98 - 99 - 100 )
C= - 8 + - 16 + -24 + ... + -200
C= - ( 8 + 16 + 24 +... + 200 )
C= - \(\frac{\left(8+200\right).25}{2}\)= - 2600
nhớ ấn đúng cho mình nhé!
Bài làm
\(M=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+...+\frac{3^2}{98.101}\)
\(M=3^2\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\right)\)
\(M=9.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(M=9\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(M=9.\left(\frac{101}{202}-\frac{2}{202}\right)\)
\(M=9.\frac{99}{202}\)
\(M=\frac{891}{202}\)
Vậy \(M=\frac{891}{202}\)
M= 3.(3/2.5+ 3/5.8.....3/98.101)
= 3.( 1/2-1/5+1/5-1/8 +....+1/98-1/101)
=3.( 1/2-1/101)
= 3.( 101/202- 2/202)
=3. 99/202
= 297/202
Vậy M= 297/202 nha bạn