không biết giải

2001 

____

1991

a: \(\left(1-\frac12\right)\times\left(1-\frac13\right)\times\ldots\times\left(1-\frac{1}{99}\right)\)

\(=\frac12\times\frac23\times\ldots\times\frac{98}{99}\)

\(=\frac{1}{99}\)

b: \(\left(x+\frac12\right)+\left(x+\frac16\right)+\left(x+\frac{1}{12}\right)+\left(x+\frac{1}{20}\right)+\cdots+\left(x+\frac{1}{90}\right)=\frac{99}{10}\)

=>\(\left(x+\frac{1}{1\times2}\right)+\left(x+\frac{1}{2\times3}\right)+\cdots+\left(x+\frac{1}{9\times10}\right)=\frac{99}{10}\)

=>\(9\times x+1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x+1-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x=\frac{99}{10}+\frac{1}{10}-1=\frac{100}{10}-1=10-1=9\)

=>x=1

\(A=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{99}\right)\)

\(A=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)

\(A=\frac{2}{99}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Dấu chấm là nhân

a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)

\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)

CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!

2/99 nha bạn mình làm rồi

ở violympic đúng ko

\(\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{99}\right)\)

\(=\frac{2}{3^{\left(1\right)}}\cdot\frac{3^{\left(1\right)}}{4^{\left(1\right)}}\cdot\frac{4^{\left(1\right)}}{5^{\left(1\right)}}\cdot...\cdot\frac{98^{\left(1\right)}}{99}\)

\(=\frac{2}{99}\)

A = 2/99 nhé !

kết  bạn với mình nha mình hết lượt  rồi

help mik với

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\)

a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)

hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)

b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow x+100=0\)

hay x=-100

Mn nhớ giải chi tiết ra nha