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AH
Akai Haruma
Giáo viên
15 tháng 10 2023

Lời giải:

$(\frac{1}{24}-\frac{5}{16}): \frac{-3}{8}+1$

$=\frac{-13}{48}.\frac{-8}{3}+1=\frac{13}{18}+1=\frac{31}{18}$

15 tháng 10 2023

=(4/96-30/96) : (-3/8) + 1 

=(-3/8) : (-3/8) + 1

=1+1

=2

7 tháng 8 2018

\(\frac{3}{5}x+\frac{1}{2}=\frac{1}{7}\)

\(\Rightarrow\frac{3}{5}x=\frac{1}{7}-\frac{1}{2}\)

\(\Rightarrow\frac{3}{5}x=\frac{-5}{14}\)

\(\Rightarrow x=\frac{-5}{14}\div\frac{3}{5}\)

\(\Rightarrow x=\frac{-25}{42}\)

8 tháng 6 2021

`1-(5 3/8+x-7 5/24):(-16 2/3)=0`
`=>1-(5+3/8+x-7-5/24):(-50/3)=0`
`=>1=(x-2-11/6):(-50/3)`
`=>1=(x-11/6):(-50/3)`
`=>x-11/6=-50/3`
`=>x=-89/6`
Vậy `x=-89/6`

8 tháng 6 2021

 

\(1-\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow1-\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right):\dfrac{50}{3}=0\)

\(\Leftrightarrow1-\left(\dfrac{129}{24}-\dfrac{173}{24}+x\right).\dfrac{3}{50}=0\)

\(\Leftrightarrow1+\dfrac{11}{6}-x.\dfrac{3}{50}=0\)

\(\Leftrightarrow\dfrac{17}{6}-x.\dfrac{3}{50}=0\)

\(\Leftrightarrow x.\dfrac{3}{50}=\dfrac{17}{6}\)

\(\Leftrightarrow x=\dfrac{425}{9}\)

-Chúc bạn học tốt-

28 tháng 6 2016

a)\(x+\frac{1}{2}:\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

   \(x+-\frac{2}{3}=-\frac{5}{4}\)

    \(x=-\frac{5}{4}-\left(-\frac{2}{3}\right)\)

    \(x=-\frac{7}{12}\)

Vậy \(x=-\frac{7}{12}\)

b)Câu b mk ko hiểu bn ghi đề sao vậy (5 3/8)

28 tháng 6 2016

a, => x + \(\frac{-2}{3}=\frac{-5}{4}\)

=>x = \(\frac{15}{8}\)

câu b mik cx làm đc nhưng mik gại ghi lắm   leuleu

Bài 1: 

1: =-5/24+16/27+3/4

=-5/24+18/24+16/27

=13/24+16/27

=117/216+128/216=245/216

2: =-1/3+1/3+6/7=6/7

3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)

4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)

6 tháng 3 2023

câu a : 

\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)

câu b : 

\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)

câu c : 

\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)

câu d : 

\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)

câu e : 

\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)

câu g : 

\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)

câu h : 

\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)

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: \(x\in\left\{0;-1;3\right\}\)

25 tháng 1 2022

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 !