2/2.4+2/4.6+2/6.8+...+2/98.100
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câu A
A :5 =1/2.4+1/4.6+1/6.8+..+1/98.100
A:5 =1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100
A:5 =1/2-1/100 =49/100
A=49/100 x5 =49/20
câu B tươg tự nha =)
Ta có:
A =5/2(1/2-1/4 + 1/4-1/6+ 1/6..........1/98-1/100)
A =5/2 (1/2 -1/100)
A =5/2 x 49/100
A = 49/20
A=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2)
A=1.2.4+2.3.4+...+50.51.4
A=4(1.2+2.3+...+50.51)
M= 1.2+2.3+...+50.51
3M=1.2.3+2.3.(4-1)+...+50.51.(52-49)
=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51
= 50.51.52
=132600
=> M=44200
=> A=4M=176800
\(B=2.4+4.6+6.8+...+98.100\)
\(B=2.\left(1.2\right)+2.\left(2.3\right)+2.\left(3.4\right)+...+2.\left(49.50\right)\)
\(B=2\left(1.2+2.3+3.4+....+49.50\right)\)
Đặt:
\(A=1.2+2.3+3.4+...+49.50\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(3A=49.50.51\)
\(A=\dfrac{49.50.51}{3}=41650\)
\(B=2A=41650.2=83300\)
6A = 2 . 4 . 6 + 4.6.6 + 6.8.6 + ... + 98 . 100 . 6 = 2.4.6 + 4.6.(8-2) + ... + 98. 100 . (102 - 96)
= 2.4.6 +4.6.8 - 2.4.6 + .... + 98.100 . 102 - 96.98.100
= 98. 100 . 102
= 999600
Suy ra A = 166600
Vậy ______________________
\(\frac{2.4+4.6+6.8+...+98.100}{1.2+2.3+3.4+...+49.50}=\frac{4.\left(1.2+2.3+3.4+...+49.50\right)}{1.2+2.3+3.4+...+49.50}=\frac{4}{1}=4\)
\(\frac{6}{2.4}+\frac{6}{4.6}+...+\frac{6}{98.100}\)
= \(\frac{6}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
= \(3.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
= \(3.\left(\frac{1}{2}-\frac{1}{100}\right)\)
= \(3.\frac{49}{100}\)
= \(\frac{147}{100}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)