a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)

\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)

a, 3/5 : -5 = -3/25

b, 24 : -6/7 = -28

c, -4/15 : 2 = -2/15

\(a.\)

\(-\dfrac{5}{9}\cdot\dfrac{12}{35}=\dfrac{\left(-5\right)\cdot12}{9\cdot35}=\dfrac{-60}{315}=-\dfrac{4}{21}\)

\(b.\)

\(\left(-\dfrac{5}{8}\right)\cdot-\dfrac{6}{55}=\dfrac{\left(-5\right)\cdot\left(-6\right)}{8\cdot55}=\dfrac{30}{440}=\dfrac{3}{44}\)

\(c.\)

\(\left(-7\right)\cdot\dfrac{2}{5}=-\dfrac{14}{5}\)

\(d.\)

\(-\dfrac{3}{8}\cdot\left(-6\right)=\dfrac{-3\cdot\left(-6\right)}{8}=\dfrac{18}{8}=\dfrac{9}{4}\)

a,-4/21 b,-3/44 c,-14/5 d,9/4

a) \(\dfrac{-5}{9}.\dfrac{12}{35}=\dfrac{-5.12}{9.35}=\dfrac{-4}{21}\) 

b) \(\dfrac{-5}{8}.\dfrac{-6}{55}=\dfrac{-5.-6}{8.55}=\dfrac{3}{44}\) 

c)\(-7.\dfrac{2}{5}=\dfrac{-7.2}{5}=\dfrac{-14}{5}\) 

d) \(\dfrac{-3}{8}.-6=\dfrac{-3.-6}{8}=\dfrac{9}{4}\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)

Câu 7:

=6

\(a.\)

\(\dfrac{3}{10}:\left(-\dfrac{2}{3}\right)=\dfrac{3}{10}\cdot\dfrac{-3}{2}=-\dfrac{9}{20}\)

\(b.\)

\(\left(-\dfrac{7}{12}\right):\left(-\dfrac{5}{6}\right)=\left(-\dfrac{7}{12}\right)\cdot\left(-\dfrac{6}{5}\right)=\dfrac{\left(-7\right)\cdot\left(-6\right)}{12\cdot5}=\dfrac{7}{10}\)

\(c.\)

\(\left(-15\right):-\dfrac{9}{10}=\left(-15\right)\cdot-\dfrac{10}{9}=\dfrac{150}{9}=\dfrac{50}{3}\)

a) \(\dfrac{3}{10}:\dfrac{-2}{3}=\dfrac{3}{10}.\dfrac{-3}{2}=\dfrac{3.-3}{10.2}=\dfrac{-9}{20}\) 

b) \(\dfrac{-7}{12}:\dfrac{-5}{6}=\dfrac{-7}{12}.\dfrac{-6}{5}=\dfrac{-7.-6}{12.5}=\dfrac{7}{10}\) 

c)\(-15:\dfrac{-9}{10}=-15.\dfrac{-10}{9}=\dfrac{-15.-10}{9}=\dfrac{50}{3}\)

a) Các phân số đảo ngược là:

\(\dfrac{5}{8}\rightarrow\dfrac{8}{5};\dfrac{3}{4}\rightarrow\dfrac{4}{3};\dfrac{1}{2}\rightarrow\dfrac{2}{1}=2\)

b) \(\dfrac{3}{7}:\dfrac{5}{8}=\dfrac{3}{7}\times\dfrac{8}{5}=\dfrac{24}{35}\)

\(\dfrac{8}{7}:\dfrac{3}{4}=\dfrac{8}{7}\times\dfrac{4}{3}=\dfrac{32}{21}\)

\(\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{1}{3}\times2=\dfrac{2\times1}{3}=\dfrac{2}{3}\)