\(a)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{22}{132}+\frac{11}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{33}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{34}{132}+\frac{1}{20}\)
\(=\frac{17}{66}+\frac{1}{20}\)
\(=\frac{203}{660}\)
\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\frac{3}{10}+\frac{1}{132}\)
\(=\frac{198}{660}+\frac{5}{660}\)
\(=\frac{203}{660}\)
1/6 + 1/12 + 1/20 + ... + 1/132
= 1/2x3 + 1/3x4 + 1/3x4 + 1/4x5 + ... + 1/11x12
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + .... + 1/11 + 1/12
= 1/2 - 1/12
= 5/12
Bài 2
= 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + 1/11x13
= 1/2 x ( 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13
= 1/2 x ( 2/3 - 2/5 + 2/5 - 2/7 + 2/7 - 2/9 + 2/9 - 2/11 + 2/11 - 2/13
= 1/2 x ( 2/3 - 2/13 )
= 1/2 x 20/39
= 10 / 39
Còn đây là câu b:
\(b,\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{10}{39}\)
\(=\frac{10}{78}=\frac{5}{39}\)
này bn nguyễn trí quân ơi.Mình thấy cách lm của bn hơi kì 1 chút .
Khi mà phân số 1/2 ở ngoài ngoặc thì trong ngoặc phần tử số luôn luôn là 1 nha.
Hãy kiểm tra lại đi nha
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\frac{3}{10}+\frac{1}{132}=\frac{203}{660}\)
\(b,\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
Mik ghi nhầm sorry
