a) 3/8 = 1/8 + 2/8 = 1/8 + 1/4

3/8 = 5/8 - 2/8 = 5/8 - 1/4

b) 5/12 = 1/12 + 4/12 = 1/12 + 1/3

5/12 = 7/12 - 2/12 = 7/12 - 1/6

c) 1/11 = -2/11 + 3/11

1/11 = 2/11 - 1/11

d) 1/4 = -2/4 + 3/4 = -1/2 + 3/4

1/4 = 5/4 - 4/4 = 5/4 -1

Câu 1:

Thay \(x=-12\) vào \(\left|x-2\right|\)

\(\Rightarrow\left|-12-2\right|=\left|-14\right|=14\)

Câu 2: Chọn phương án A.

Câu 3:

\(\left|-120\right|+\left|20\right|=120+20=140\)

Câu `1`

` |x + 2|`

mà `x=-12`

`->  |-12 + 2|= |-10|=10`

`->B`

Câu `2`

`->A`

Câu `3`

`A = |-120| + |20|`

`= 120 +20`

`=140`

a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

   b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)

   c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)

   d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)

b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

a.=\(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{61}{4}:\dfrac{7}{12}=\dfrac{471}{14}-\dfrac{183}{7}=\dfrac{15}{2}\)

b.=\(\dfrac{2}{15}-\dfrac{2}{3}+\dfrac{1}{5}=-\dfrac{1}{3}\)

c.\(\left(\dfrac{10}{3}+2.5\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}=\dfrac{35}{6}:\left(-\dfrac{31}{30}\right)-\dfrac{11}{31}=-\dfrac{175}{31}-\dfrac{11}{31}=-6\)

d.\(\left[6+\dfrac{1}{8}-\dfrac{1}{2}\right]:\dfrac{3}{12}=\dfrac{45}{8}:\dfrac{3}{12}=\dfrac{45}{2}\)