\(2P=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)
\(2P=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\)
\(2P=\frac{1}{2}-\frac{1}{100}\)
=> P =\(\frac{49}{100}:2=\frac{49}{100}\cdot\frac{1}{2}=\frac{49}{200}\)
\(2P=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)
\(2P=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\)
\(2P=\frac{1}{2}-\frac{1}{100}\)
=> P =\(\frac{49}{100}:2=\frac{49}{100}\cdot\frac{1}{2}=\frac{49}{200}\)
Tính nhanh:(1/2.4)+(1/4.6)+(1/6.8)+.....+(1/98.100)=
1/1.3-1/2.4+1/3.5-1/4.6+...+1/97.99-1/98.100 = ?
Tính nhanh:
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{96.98}+\dfrac{1}{98.100}\)
1.tính
\(d=2.4+4.6+........+98.100\)
giúp mình bài này với nhé!!!!
1/2.4 + 1/4.6 + 1/6.8 + ... + 1/98.100
có bn nào lm giùm mik vs
chứng minh rằng:1/1.3 + 1/2.4 + 1/3.5 + 1/4.6 +....+ 1/97.99 + 1/98.100 < 3/4
chứng minh rằng 1^1.3 + 1^2.4 + 1^3.5 + 1^4.6 +...+ 1^97.99+ 1^98.100 < 3^4
Tính A
A=2.4 + 4.6 + ..... + 98.100