a) (26-6)*(-4)+31*(-7-13)

=-80+-620

=-700

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a: \(\left(5+\frac15-\frac29\right)-\left(2-\frac{1}{23}-\frac{3}{35}+\frac56\right)-\left(8+\frac27-\frac{1}{18}\right)\)

\(=5+\frac15-\frac29-2+\frac{1}{23}+\frac{3}{35}-\frac56-8-\frac27+\frac{1}{18}\)

\(=\left(5-2-8\right)+\left(\frac15+\frac{3}{35}-\frac27\right)+\left(-\frac29-\frac56+\frac{1}{18}\right)+\frac{1}{23}\)

\(=\left(-5\right)+\left(\frac{7}{35}+\frac{3}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)

\(=-5+\left(-\frac{18}{18}\right)+\frac{1}{23}=-6+\frac{1}{23}=-\frac{138}{23}+\frac{1}{23}=-\frac{137}{23}\)

c: \(-\frac57-\left(-\frac{5}{67}\right)+\frac{13}{10}+\frac12+\left(-\frac16\right)+1\frac{3}{14}-\left(-\frac25\right)\)

\(=-\frac57+\frac{5}{67}+\frac{13}{10}+\frac12-\frac16+\frac{17}{14}+\frac25\)

\(=\left(-\frac57+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac25-\frac16\right)+\frac{5}{67}\)

\(=\left(-\frac{10}{14}+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac{4}{10}-\frac16\right)+\frac{5}{67}\)

\(=\left(\frac{7}{14}+\frac12\right)+\left(\frac{17}{10}-\frac16\right)+\frac{5}{67}=1+\frac{5}{67}+\frac{51}{30}-\frac{5}{30}\)

\(=\frac{72}{67}+\frac{46}{30}=\frac{72}{67}+\frac{23}{15}=\frac{2621}{1005}\)

d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)

\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)

\(=\frac35:\left(-\frac{7}{30}\right)+\frac35:\left(-\frac{21}{15}\right)\)

\(=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\cdot\left(-\frac{30}{7}-\frac57\right)\)

\(=\frac35\cdot\left(-\frac{35}{7}\right)=\frac35\cdot\left(-5\right)=-3\)

=???

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cái nịt

Bài 1: 

1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)

\(=-12-18\)

=-30

2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)

\(=36-2020+2019-136-27\)

\(=1-100-27\)

\(=-126\)

3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)

\(=144-97-244+197\)

\(=-100+100=0\)

4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)

\(=-24\cdot13+24\cdot3\)

\(=24\cdot\left(-13+3\right)\)

\(=24\cdot\left(-10\right)=-240\)

5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)

\(=54+55+56+57+58-64-65-66-67-68\)

\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)

\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)

=-50

6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)

\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)

\(=-24\cdot5+16\cdot5\)

\(=5\cdot\left(-24+16\right)\)

\(=-5\cdot8=-40\)

7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)

\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)

\(=47\cdot50-23\cdot50\)

\(=50\cdot\left(47-23\right)\)

\(=50\cdot24=1200\)

8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)

\(=-31\cdot\left(47+52+1\right)\)

\(=-31\cdot100=-3100\)

Bài 2: 

1) Ta có: \(-17-\left(2x-5\right)=-6\)

\(\Leftrightarrow-17-2x+5+6=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow-2x=6\)

hay x=-3

Vậy: x=-3

2) Ta có: \(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow10-8+6x+4=0\)

\(\Leftrightarrow6x+6=0\)

\(\Leftrightarrow6x=-6\)

hay x=-1

Vậy: x=-1

3) Ta có: \(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow-12-3x+21+18=0\)

\(\Leftrightarrow-3x+27=0\)

\(\Leftrightarrow-3x=-27\)

hay x=9

Vậy: x=9

4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)

\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)

\(\Leftrightarrow-2x-3=-3\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: x=0

5) Ta có: x(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3\right\}\)

6) Ta có: (x-2)(x+4)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-4\right\}\)

7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-1;3\right\}\)

Bài 1: 

1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)

=−12−18=−12−18

=-30

2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27

=36−2020+2019−136−27=36−2020+2019−136−27

=1−100−27=1−100−27

=−126

Tớ chcs cậu học thật giỏi nha !

A) – 4;            B) – 1;

C) – 3;            D) – 5

73+36+968+914+3032

=73+(36+914)+(968+3032)

=73+950+4000

=5023

1) ( mk khong viet lai de dau nhe! )

a)= 73 + ( 36 + 914 ) + ( 968 + 3032 )

= 73 + 950 + 4000

= 1023 + 4000

= 5023

b)=314 x ( 67 + 16) + 659 x 83

= 314 x 83 + 659 x 83

= ( 314 + 659 ) x 83

= 973 x 83

= 80759