K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 2 2022

\(x=-21/13+7/13=-14/13\) nhé

20 tháng 2 2022

sao dấu trừ thành cộng 

16 tháng 4 2022

= -103/273

= 0

= -5/7

16 tháng 4 2022

5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)

b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)

c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)

6 tháng 2 2023

d) 

` (-3)/7 + 5/13 + (-4)/7 =   -3/7 + (-4/7 )  + 5/13 = -1 + 5/13 = -8/13`

e2) 

`-5/21 + (-2)/21 + (-8)/4 = -7/21 + -8/4 = -7/3`

25 tháng 2 2023

\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)

\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)

\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)

\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)

\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)

\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)

1 tháng 8 2017

\(\dfrac{-8}{13}+\dfrac{-7}{17}+\dfrac{21}{13}\le x\le\dfrac{-9}{14}+3+\dfrac{5}{-14}\)

=> \(\dfrac{10}{17}\le x\le2\)

=> \(\dfrac{10}{17}\le\dfrac{17x}{17}\le\dfrac{34}{17}\)

=> 10 \(\le17x\le34\)
=> x = 1; 2 (thỏa mãn)
@Khánh Linh

20 tháng 7 2018

a, 4,5 - 2x . 147 = 1114

2x . 147 = 4,5 - 1114

2x . 147 = -1109,5

2x = -1109,5 / 147

2x = \(-\dfrac{317}{42}\)

x = \(-\dfrac{317}{42}:2\)

x = \(-\dfrac{317}{84}\)

b, ( x + 14 - 13 ) : ( 2 + 16 - 14 ) = 746

( x + 1 ) : 4 = 746

x + 1 = 746 . 4

x + 1 = 2984

x = 2984 - 1

x = 2983

c, \(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{7}{10}\)

\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\)

\(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{1}{6}\)

\(\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\)

\(\dfrac{13}{21}+x=\dfrac{2}{7}\)

\(x=\dfrac{2}{7}-\dfrac{13}{21}\)

\(x=-\dfrac{1}{3}\)

Ta có: \(A=\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)

\(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)

\(=1-1+\dfrac{-5}{7}\)

\(=\dfrac{-5}{7}\)

Có 2 dấu = hả bạn

`# \text {DNamNgV}`

`16/21 + 6/7`

`= 16/21 + 18/21`

`= 34/21`

__

`15/22 \times 11/35`

`= (15 \times 11)/(22 \times 35)`

`= (5 \times 3 \times 11)/(2 \times 11 \times 5 \times 7)`

`= (3 \times 1)/(2 \times 7)`

`= 3/14`

___

`8/11 \div 5/22`

`= 8/11 \times 22/5`

`= (8 \times 22)/(11 \times 5)`

`= (8 \times 11 \times 2)/(11 \times 5)`

`= (8 \times 2)/5`

`= 16/5`

___

`9/13 \div 27/39`

`= 9/13 \times 39/27`

`= 9/13 \times 13/9`

`= 1`

4 tháng 8 2018

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

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

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)