`28 xx 7,32 - 7,32 : 0,125`

`= 28 xx 7,32 - 7,32 xx8`

`= 7,32 xx(28-8)`

`= 7,32 xx20`

`=146,4`

`1 - 2 + 3 - 4 + 5 - 6 +...+ 2021 - 2022`

`= (1 - 2) + (3 - 4) + (5 - 6) +...+ (2021 - 2022)`

`= (-1) + (-1) + (-1) + ... + (-1) `  [có `2022 : 2 = 1011` nhóm]

`= (-1) xx 1011 = -1011`

a) \(\left(29,42+20,58\right)-\left(34,23+25,77\right)\)

\(=50-60=-10\)

b) \(-\left(212,49+87,51\right)+99,9\)

\(=-300+99,9=-200,1\)

A)29.42+20.58−34.23+(−25.77)29.42+20.58−34.23+(−25.77) 

=29.42+20.58−34.23−25.77

=29.42+20.58−34.23−25.77

=(29.42+20.58)−(34.23+25.77)

=(29.42+20.58)−(34.23+25.77)

=50−60=50−60

=−10=−10

B).(−212.49)−(87.51−99.9)(−212.49)−(87.51−99.9) 

=−212.49−87.51+99.9

=−212.49−87.51+99.9

=−(212.49+87.51)+99.9

=−(212.49+87.51)+99.9

=−300+99.9=−300+99.9

=−(300−99.9)=−(300−99.9)

=−200.1

a) \(3119\cdot121-3119\cdot11\cdot11\\ =3119\left(121-11\cdot11\right)\\ =3119\left(121-121\right)\\ =3119\cdot0=0\)

b) \(24\cdot\left(123+87\right)+\left(87+123\right)\cdot76\\ =\left(123+87\right)\left(24+76\right)\\ =210\cdot100=21000\)

a: \(232-\left(581+132-331\right)\)

\(=232-581-132+331\)

\(=\left(232-132\right)+\left(-581+331\right)\)

\(=100-250\)

=-150

b: \(\left[12+\left(-57\right)\right]-\left[-57-\left(-12\right)\right]\)

\(=12+\left(-57\right)+57+\left(-12\right)\)

\(=\left(12-12\right)+\left(57-57\right)\)

=0+0

=0

a)\(=12\times3+4\times12+12\times3=12\times\left(3+4+3\right)=12\times10=120\)

b)\(=25\times7-25\times2-3\times25-2\times25=25\times\left(7-2-3-2\right)=0\)

A) 12x3+4x12+36

=12x3+4x12+12x3

=12x(3+4+3)

=120

\(a,\dfrac{-19}{34}-\left(\dfrac{31}{51}-\dfrac{54}{35}\right)=\dfrac{-19}{34}-\left(-\dfrac{1669}{1785}\right)=\dfrac{79}{210}\\ b,\dfrac{-2}{5}.\dfrac{4}{15}+\left(-0,4\right).\dfrac{26}{15}=-\dfrac{8}{75}-\dfrac{52}{75}=-\dfrac{4}{5}\)

Tích giúp mình nha

b) Ta có: \(86\cdot15+150\cdot1.4\)

\(=86\cdot15+14\cdot15\)

\(=15\left(86+14\right)\)

=1500

c) Ta có: \(93\cdot32+14\cdot16\)

\(=186\cdot16+14\cdot16\)

\(=16\left(186+14\right)\)

\(=16\cdot200=3200\)

\(\dfrac{1}{2}\cdot1\cdot\dfrac{1}{3}\cdot10\cdot\dfrac{7}{35}\cdot\dfrac{3}{4}\)

\(=\dfrac{1}{2}\cdot10\cdot\dfrac{1}{5}\cdot1\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\)

\(=\dfrac{10}{10}\cdot\dfrac{1}{4}=\dfrac{1}{4}\)

a: =53(7x8-87)+538

=-1105

c.=615