\(\dfrac{3}{13}-\left(\dfrac{4}{7}+\dfrac{3}{13}\right)=\dfrac{3}{13}-\dfrac{4}{7}-\dfrac{3}{13}=\left(\dfrac{3}{13}-\dfrac{3}{13}\right)-\dfrac{4}{7}=0-\dfrac{4}{7}=-\dfrac{4}{7}\)

= 3/13 - 4/7 - 3/13
= (3/13 - 3/13) - 4/7
=0- 4/7
= -4/7

1: \(\dfrac{-5}{12}\cdot\dfrac{2}{11}+\dfrac{-5}{12}\cdot\dfrac{9}{11}+\dfrac{5}{12}\)

\(=-\dfrac{5}{12}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{5}{12}\)

\(=-\dfrac{5}{12}+\dfrac{5}{12}=0\)

2: \(\dfrac{-3}{5}:\dfrac{7}{5}+\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)

\(=\left(-\dfrac{3}{5}+\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)

\(=0\cdot\dfrac{5}{7}+\dfrac{13}{5}=\dfrac{13}{5}\)

3: \(\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+\left(-2022\right)^0\)

\(=\dfrac{-3}{7}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+1\)

\(=-\dfrac{3}{7}+1=\dfrac{4}{7}\)

4: \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\dfrac{-1}{9}\)

\(=0,75-\dfrac{7}{3}-0,75+9\cdot\dfrac{-1}{9}\)

\(=-\dfrac{7}{3}-1=-\dfrac{10}{3}\)

5: \(2\dfrac{6}{7}\cdot\left[\left(-\dfrac{7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2\right]\)

\(=\dfrac{20}{7}\cdot\left[-\dfrac{7}{5}-\dfrac{3}{2}\cdot\dfrac{4}{5}+\dfrac{9}{4}\right]\)

\(=\dfrac{20}{7}\left(-\dfrac{7}{5}-\dfrac{3}{10}+\dfrac{9}{4}\right)\)

\(=\dfrac{20}{7}\cdot\dfrac{-28-6+45}{20}\)

\(=\dfrac{45-34}{7}=\dfrac{11}{7}\)

6: \(\dfrac{2}{7}+\dfrac{5}{7}\left(\dfrac{3}{5}-0,25\right)\cdot\left(-2\right)^2+35\%\)

\(=\dfrac{2}{7}+\dfrac{5}{7}\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot4+\dfrac{7}{20}\)

\(=\dfrac{2}{7}+\dfrac{7}{20}+\dfrac{20}{7}\cdot\dfrac{7}{20}\)

\(=\dfrac{89}{140}+1=\dfrac{239}{140}\)

7: \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{11}{20}+25\%\right):1\dfrac{2}{5}\)

\(=\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{5}{20}\right):\dfrac{7}{5}\)

\(=\dfrac{21}{15}-\dfrac{16}{20}\cdot\dfrac{5}{7}\)

\(=\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{49-20}{35}=\dfrac{29}{35}\)

8: \(\left(-2,4+\dfrac{1}{3}\right):3\dfrac{1}{10}+75\%:1\dfrac{1}{2}\)

\(=\left(-\dfrac{12}{5}+\dfrac{1}{3}\right):\dfrac{31}{10}+\dfrac{3}{4}:\dfrac{3}{2}\)

\(=\dfrac{-31}{15}\cdot\dfrac{10}{31}+\dfrac{1}{2}\)

\(=-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{-4+3}{6}=\dfrac{-1}{6}\)

\(\left(-1,5\right)\cdot\dfrac{-2}{3}+\left(2,5-\dfrac{3}{4}\right):1\dfrac{3}{4}\)

\(=\dfrac{-3}{2}\cdot\dfrac{-2}{3}+\left(2,5-0,75\right):1,75\)

=1+1

=2

Tỉ số phần trăm giữa số học sinh nữ và số học sinh nam là:

\(\dfrac{1}{2}=50\%\)

a.

\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1.3}{2.3}-\dfrac{1.2}{2.3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)

b.

\(\dfrac{12}{17}+\dfrac{5}{11}+\dfrac{5}{17}+\dfrac{-16}{11}=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{5}{11}+\dfrac{-16}{11}\right)\)

\(=\dfrac{17}{17}+\dfrac{-11}{11}=1-1=0\)

Gọi \(d=ƯC\left(n+1;n+2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\left(n+1\right)⋮d\\\left(n+2\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left(n+2\right)-\left(n+1\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=\pm1\)

\(\Rightarrow\dfrac{n+1}{n+2}\) là phân số tối giản với mọi \(n\in Z\)

CM:A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\) < 1

   \(\dfrac{1}{2^2}\) = \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}-\dfrac{1}{2}\)

   \(\dfrac{1}{3^2}\) = \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)

    \(\dfrac{1}{4^2}\) = \(\dfrac{1}{4.4}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)

   \(\dfrac{1}{100^2}\) = \(\dfrac{1}{100.100}\) < \(\dfrac{1}{99.100}\) = \(\dfrac{1}{99}-\dfrac{1}{100}\)

Cộng vế với vế ta có:

    \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\)  = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\) < 1 (đpcm)

(1/2)ˣ = 1/8

(1/2)ˣ = (1/2)³

x = 3

\(\dfrac{2}{3}x\) - \(\dfrac{1}{2}x\) = \(\dfrac{5}{12}\)

\(\dfrac{1}{6}x\)         = \(\dfrac{5}{12}\)

   \(x\)         = \(\dfrac{5}{12}\) : \(\dfrac{1}{6}\)

    \(x\)       = \(\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)