a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)

b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)

c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)

a: \(\left(\frac{5}{12}-\frac34\right)\cdot\frac{-6}{5}+\frac{-7}{6}\)

\(=\left(\frac{5}{12}-\frac{9}{12}\right)\cdot\frac{-6}{5}+\frac{-7}{6}\)

\(=\frac{-4}{12}\cdot\frac{-6}{5}+\frac{-7}{6}=\frac25+\frac{-7}{6}\)

\(=\frac{12-35}{30}=-\frac{23}{30}\)

b: \(\left(5\frac12-5\right)^2+\frac{-1}{6}:\frac53\)

\(=\left(\frac12\right)^2+\frac{-1}{6}\cdot\frac35\)

\(=\frac14-\frac{1}{10}=\frac{5}{20}-\frac{2}{20}=\frac{3}{20}\)

c: \(75\%-1\frac12+0,5:\frac{5}{12}\)

\(=\frac34-\frac32+0,5\cdot\frac{12}{5}\)

\(=-\frac34+\frac65=\frac{-15+24}{20}=\frac{9}{20}\)

a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)

b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)

c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)

d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)

Lời giải:

a. 

$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$

$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$

$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$

b.

$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$

$3:x=\frac{1}{4}$

$x=3: \frac{1}{4}=12$

`a)1/2+[-1]/[-3]-5/12 < 2x < 12/[-31]+136/31`

`186/372+124/372-155/372 < [744x]/372 < [-144]/372+1632/372`

`186+124-155 < 744x < -144+1632`

`155 < 744x < 1488`

`155:744 < 744x:744 < 1488:744`

`5/24 < x < 2`

Vậy `5/24 < x < 2`

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`b)[-2]/5 < x/15 < 1/6`

`[-12]/30 < [2x]/30 < 5/30`

`-12 < 2x < 5`

`-12:2 < 2x:2 < 5:2`

`-6 < x < 5/2`

Vậy `-6 < x < 5/2`

Giải:

a) x - \(\dfrac{9}{25}\)= \(\dfrac{16}{25}\)

x = \(\dfrac{16}{25}\)+\(\dfrac{9}{25}\)
x = \(\dfrac{25}{25}\)

x = 1

b) \(\dfrac{-12}{30}\)<\(\dfrac{x}{30}\)<\(\dfrac{5}{30}\)

=> x có thể bằng \(\dfrac{-11}{30}\) đến \(\dfrac{4}{30}\)
=> x bằng -5; -4; -3; -2; -1;0;1;2

\(\frac12+\frac{1}{12}+\frac{1}{30}+\cdots+\frac{1}{9900}\)

\(=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{99\cdot100}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac12+\cdots+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{100}\right)\)
\(=1+\frac12+\cdots+\frac{1}{99}+\frac{1}{100}-1-\frac12-\cdots-\frac{1}{50}\)

\(=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)

Ta có: \(A=\left(\frac12+\frac{1}{12}+\frac{1}{30}+\cdots+\frac{1}{9900}\right):\left(\frac{-6}{51}-\frac{6}{52}-\cdots-\frac{6}{100}\right)\)

\(=\frac{\frac{1}{51}+\frac{1}{52}+\cdots+100}{-6\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\right)}=\frac{-1}{6}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(A=1-\dfrac{1}{8}=\dfrac{7}{8}\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(A=\dfrac{367}{420}\approx0,87\)

Mình đang cần gấp!☹

Đ V T H

Cặp 1: \(\frac12+\frac{1}{12}=\frac{7}{12}\)

Cặp 2: \(2+\frac16=\frac32+\frac23\)

Cặp 3: \(\frac{1}{20}=\frac14-\frac15\)

Cặp 4: \(1-\frac19=\frac13+\frac59\)

Cặp 5: \(\frac{1}{15}+\frac{2}{15}=\frac15\)

Cặp 6: \(\frac12+\frac23=1+\frac16\)

Cặp 7: \(\frac{4}{12}=1-\frac23\)

Cặp 8: \(\frac{9}{14}+\frac{1}{14}=\frac57\)

Cặp 9: \(\frac13-\frac29=1-\frac89\)