\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot...\cdot1\frac{1}{99}\)

\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{100}{99}\)

\(=\frac{2.2\cdot3.3\cdot4.4\cdot5.5\cdot...\cdot10.10}{1.3\cdot2.4\cdot3.5\cdot4.6\cdot...\cdot9.11}\)

\(=\frac{2.10}{1.11}=\frac{20}{11}\)

"." = nhân

\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}+\frac{1}{99}\)

\(=2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)

\(=2-\frac{1}{99}\)

\(=\frac{197}{99}\)

1+1/3+1/5...+1/97+1/99=(1+1/99) + (1/3+1/97) + (1/5+1/95)....+(1/49+1/51) 
= 100/1.99 + 100/3.97 + 100/5.95 +.....=100.(1/1.99 + 1/3.97 + 1/5.95 +.....) 
Mau so: 
1/1.99 + 1/3.97 +1/5.95....+1/95.5+ 1/97.3 +1/99.1=2/1.99 +2/3.97 +2/5.95+..... 
=2.(1/1.99 + 1/3.97 + 1/5.95 +.....) 
=>A=(100.(1/1.99 + 1/3.97 + 1/5.95 +.....)) : (2.(1/1.99 + 1/3.97 + 1/5.95 +.....))=50

chuẩn luôn , tích nha

Thanks nhìu  ^_^

đúng ko hoài mình cho bạn nhé

ai tích mình mình tích lại

\(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)

=> 1=1

cảm ơn bạn nhé !

98/303 tích nha mình giải cho

Ta có : 

\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2-\frac{1}{64}=0\)

\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1}{64}\)

\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1^2}{8^2}\)

\(\Leftrightarrow\)\(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)

\(\Leftrightarrow\)\(\frac{3}{4}x-\frac{1}{2}=\frac{1}{8}\)

\(\Leftrightarrow\)\(\frac{3}{4}x=\frac{1}{8}+\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{3}{4}x=\frac{5}{8}\)

\(\Leftrightarrow\)\(x=\frac{5}{8}:\frac{3}{4}\)

\(\Leftrightarrow\)\(x=\frac{5}{8}.\frac{4}{3}\)

\(\Leftrightarrow\)\(x=\frac{5}{2}.\frac{1}{3}\)

\(\Leftrightarrow\)\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\)

Chúc bạn học tốt ~ 

\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2-\frac{1}{64}=0\)

\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2=0+\frac{1}{64}=\frac{1}{64}\)

\(\left(\frac{3}{4}.x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)

=>\(\frac{3}{4}.x-\frac{1}{2}=\frac{1}{8}\)

\(\frac{3}{4}.x=\frac{1}{8}+\frac{1}{2}\)

\(\frac{3}{4}.x=\frac{5}{8}\)

\(x=\frac{5}{8}:\frac{3}{4}\)

\(x=\frac{5}{6}\)

=> \(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\frac{1}{64}\)

Mà \(\frac{1}{64}=\left(\frac{1}{8}\right)^2=\left(\frac{-1}{8}\right)^2\)

+ Với \(\left(\frac{3}{4}x-\frac{1}{2}\right)^2=\left(\frac{1}{8}\right)^2\)

=> \(\frac{3}{4}x-\frac{1}{2}=\frac{1}{8}\)

=> \(\frac{3}{4}x=\frac{1}{8}+\frac{1}{2}\)

=> \(\frac{3}{4}x=\frac{2}{16}+\frac{8}{16}\)

=> \(\frac{3}{4}x=\frac{5}{8}\)

=> x = \(\frac{5}{8}:\frac{3}{4}\)

=> x = \(\frac{5}{6}\)

+ Với

=>

=>

=>

=>

=> x = \(\frac{1}{2}\)

Vậy x = 1/2 hoặc x = 5/6

a, \(\left|x+2\right|-\left|x+7\right|=0\Rightarrow\left|x+2\right|=\left|x+7\right|\Rightarrow\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}\Rightarrow\orbr{\begin{cases}0=5\left(loại\right)\\2x=-9\end{cases}\Rightarrow}x=\frac{-9}{2}}\)

b, - Nếu \(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\), ta có: 2x - 1 = 2x - 1 => 2x = 2x (thỏa mãn với mọi x)

- Nếu 2x - 1 < 0 => \(x< \frac{1}{2}\), ta có: 2x - 1 = 1 - 2x => 4x = 2 => x = \(\frac{1}{2}\) (không thỏa mãn điều kiện)

Vậy \(x\ge\frac{1}{2}\)

c,d tương tự b

e, tương tự a