Bài 3:
a: \(A=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\frac{5^2}{4\cdot6}\)
\(=\frac{2\cdot3\cdot4\cdot5}{1\cdot2\cdot3\cdot4}\cdot\frac{2\cdot3\cdot4\cdot5}{3\cdot4\cdot5\cdot6}=\frac51\cdot\frac26=\frac53\)
b: \(B=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\ldots\cdot\left(1-\frac{1}{100^2}\right)\)
\(=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{100}\right)\left(1+\frac12\right)\left(1+\frac13\right)\cdot\ldots\cdot\left(1+\frac{1}{100}\right)\)
\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}\)
Bài 2:
a: \(\left(2\frac45x-50\right):\frac23=51\)
=>\(2,8x-50=51\cdot\frac23=34\)
=>2,8x=50+34=84
=>x=30
b: \(x\cdot1\frac34+\left(-\frac76\right)\cdot x-1\frac23=\frac{5}{12}\)
=>\(x\left(\frac74-\frac76\right)=\frac{5}{12}+\frac53=\frac{5}{12}+\frac{20}{12}=\frac{25}{12}\)
=>\(x\left(\frac{21}{12}-\frac{14}{12}\right)=\frac{25}{12}\)
=>\(x\cdot\frac{7}{12}=\frac{25}{12}\)
=>\(x=\frac{25}{7}\)
c: \(\left(x+\frac12\right)\left(\frac23-2x\right)=0\)
=>\(\left[\begin{array}{l}x+\frac12=0\\ \frac23-2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ 2x=\frac23\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=\frac13\end{array}\right.\)
