Question

Tính

S = 1/1.2+1/2.3+1/3.4+1/4.5

P= 1/1.3+1/3.5+1/5.7+1/7.9

Answer 1

s = 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5

S=1 + (-1/2 +1/2)+...+(-1/4 + 1/4 ) +-1/5

S = 1 + 0 +0 +...+ 0 +-1/5

S= 1 + -1/5

S = 4/5

Answer 2

S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5

S=1-1/5

S=4/5.

P=1/1.3+1/3.5+1/5.7+1/7.9

2P=2/1.3+2/3.5+2/5.7+2/7.9

2P=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

2P=1-1/9=8/9

P=8/9:2

P=4/9.

Chac chan dung nha ban.k cho minh nhe

Answer 3

S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5 =1/1-1/5 nhé  P=1/2+(2/1.3+2/3.5+2/5.7+2/7.9)=1/2+(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9).=1/2+(1/1-1/9)

Answer 4

k cho minh rui minh se giai ngay

Answer 5

k cho minh nhe

Answer 6

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(S=1-\frac{1}{5}\)

\(S=\frac{4}{5}\)

\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(P=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)

\(P=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)

\(P=\frac{1}{2}.\frac{8}{9}\)

\(P=\frac{8}{18}\)

Answer 7

\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)=\(S=1-\frac{1}{5}=\frac{4}{5}\)

Vậy....

\(P=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}=2\cdot P=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)\(\Rightarrow2\cdot P=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\Rightarrow2\cdot P=1-\frac{1}{9}=\frac{8}{9}\Rightarrow P=\frac{8}{9}\div2\)\(\Rightarrow P=\frac{8}{9}\cdot\frac{1}{2}\Rightarrow P=\frac{4}{9}\)

Vậy ....

Answer 8

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(=1-\frac{1}{5}=\frac{4}{5}\)

\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{9}\right)=\frac{4}{9}\)