\(=\dfrac{6}{119}\)

\(\dfrac{13}{17}-\dfrac{23}{31}+\dfrac{4}{17}-\dfrac{8}{31}\)

\(=\left(\dfrac{13}{17}+\dfrac{4}{17}\right)-\left(\dfrac{23}{31}+\dfrac{8}{31}\right)\)

\(=\dfrac{17}{17}-\dfrac{31}{31}\)

\(=1-1\)

\(=0\)

13/17 - 23/31 + 4/17- 8/31

= (13/17 + 4/17) - (23/31 - 8/31)

= 1 - 15/31

= 16/31

\(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)\) =\(\frac{31}{23}-\frac{417}{736}=\frac{25}{32}\)

\(\frac{25}{32}\)

31/23-(7/32+8/23) =31/23-417/736=25/32

31/23-(7/32+8/23)

=31/23-7/32-8/23

=(31/23-8/23)-7/32

=23/23-7/32

=1-7/32=32/32-7/32=25/32

31 / 23 - ( 7 / 23 + 8 / 23 ) = 31 / 23 - 7 / 23 - 8 / 23 

                                            = ( 31 / 23 + ( -8 ) / 23 ) + ( - 7 ) / 23

                                            =   23 / 23 + ( -7 ) / 23

                                            =  16 / 23

\(\frac{19}{15}+\frac{31}{23}-\frac{4}{15}-\frac{8}{23}=\frac{19-4}{15}+\frac{31-8}{23}=\frac{15}{15}+\frac{23}{23}=1+1=2\)

\(\frac{19}{15}+\frac{31}{23}-\frac{4}{15}-\frac{8}{23}\)

\(=\frac{19-4}{15}+\frac{31-8}{23}\)

\(=\frac{15}{15}+\frac{23}{23}\)

\(=1+1=2\)

a: =-12/20+5/20-6/20

=-13/20

b: =10/50-45/50-14/50

=-49/50

c: =31/23-7/32-8/23=1-7/32=25/32

`a)-3/5 + 1/4 + -3/10= (-12)/20 + 5/20 + (-6)/20 = -13/20` 

`b)1/5 + -9/10 + (-7/25)= 10/50 + (-45)/50 + (-14)/50 = (-49)/50`

`c)31/23 - ( 7/32 + 8/23 )=  31/23 - 417/736 =992/736 - 417/736  =  575/736=25/32  `

\(a\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)

\(=\frac{31}{23}-\frac{7}{23}-\frac{8}{23}\)

\(=\frac{16}{23}\)

Hot

tốt

!

a)\(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)=\frac{31}{23}-\frac{8}{23}+\frac{7}{23}=1+\frac{7}{23}=1\frac{7}{23}\)

b)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)=\left(\frac{13}{41}+\frac{28}{41}\right)+\left(\frac{79}{67}-\frac{12}{67}\right)-\frac{1}{3}=1+1-\frac{1}{3}\)

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

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) 2016 - 100.(x+31)=2⁷:2³ 

<=> 2016 - 100.(x+31) = 2⁴ = 16

<=> 100.(x+31) = 2016 - 16 = 2000

<=> x+31 = 2000:100 = 20

<=> x = 20 - 31 = -11

b) (x+8)³ = 125

<=> (x+8)³ = 5³ 

<=> x+8 = 5

<=> x = 5 - 8

<=> x = -3