\(\frac{3}{5}\)+\(\frac{-1}{3}\)

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30 tháng 9 2020

a, \(\frac{3}{5}+\frac{-1}{3}\)

=\(\frac{9}{15}+\frac{-5}{15}\)

=\(\frac{4}{15}\)

b,\(\frac{-2}{13}+\frac{-11}{26}\)

=\(\frac{-4}{26}+\frac{-11}{26}\)

=\(\frac{-15}{26}\)

c, \(-2+\frac{-5}{8}\)

=\(\frac{-16}{8}+\frac{-5}{8}\)

=\(\frac{-21}{8}\)

d,\(\frac{13}{30}-\frac{1}{5}\)

=\(\frac{13}{30}-\frac{6}{30}\)

=\(\frac{7}{30}\)

e,\(\frac{2}{21}-\frac{-1}{28}\)
=\(\frac{56}{588}-\frac{-21}{588}\)
=\(\frac{77}{588}=\frac{11}{84}\)

bài dễ mà <3

4 tháng 9 2020

              Bài làm :

a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)

b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)

c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)

d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)

e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)

20 tháng 9 2019

e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)

\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)

\(=\frac{1}{7}.\left(-2\right)\)

\(=-\frac{2}{7}.\)

Chúc bạn học tốt!

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)

b) \(\left(2x-1\right).\left(2x+3\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)

\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)

d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

28 tháng 7 2019

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

=> \(x:\frac{1}{45}=\frac{1}{2}\)

=> \(x=\frac{1}{2}.\frac{1}{45}\)

=> \(x=\frac{1}{90}\)

Vậy \(x=\frac{1}{90}.\)

b) \(\left(2x-1\right).\left(2x+3\right)=0\)

=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)

Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.

Chúc bạn học tốt!

16 tháng 9 2019

e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)

\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)

\(=\frac{11}{42}:\frac{1}{28}-8\)

\(=\frac{22}{3}-8\)

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

16 tháng 9 2019

e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)

\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)

\(=\frac{11}{42}:\frac{1}{28}-8\)

\(=\frac{22}{3}-8\)

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

23 tháng 7 2017

B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7  + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15

= 0 -0-0-0-0+7/9 +13/15

= 74/45

25 tháng 8 2018

b, Nhóm các cặp trái dấu vào với nhau thì hết cuối cùng còn 13/15

c,\(\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...+\frac{1}{2}-\frac{1}{3}+1\)

\(\frac{1}{6}+1\)= 7/6

24 tháng 4 2020

1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)

\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)

( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))

2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)

\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)

Vậy \(x=\frac{-92}{105}\)

b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)

Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)

29 tháng 3 2020
https://i.imgur.com/gAuTlsH.jpg
29 tháng 3 2020

d) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)

\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)

\(=-3+\frac{3}{2}+1-5+\frac{6}{5}-\frac{3}{2}\)

\(=-7+\frac{6}{5}=-\frac{29}{5}\)