( 75%+\(\dfrac{7}{3}\)): ( \(\dfrac{2}{9}\) -2\(\dfrac{5}{2}\))
1/ (\(\left(-\dfrac{2}{3}\right)\)\(^2\) x \(\dfrac{-9}{8}\) - 25% x \(\dfrac{-16}{5}\)
2/ -1\(\dfrac{2}{5}\) x 75% + \(\dfrac{-7}{5}\) x 25%
3/ -2\(\dfrac{3}{7}\) x (-125%) + \(\dfrac{-17}{7}\) x 25%
4/ (-2)\(^3\) x (\(\dfrac{3}{4}\) x 0.25) : (2\(\dfrac{1}{4}\) - 1\(\dfrac{1}{6}\))
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
a)\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{1}{5}=9\dfrac{5}{7}\)
b)75%x-x=\(-1\dfrac{3}{4}\)
a)\(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{1}{5}=9\dfrac{5}{7}\)
\(\left(x-\dfrac{1}{2}\right).3+\dfrac{1}{5}=\dfrac{68}{7}\)
\(3x-\dfrac{3}{2}+\dfrac{1}{5}=\dfrac{68}{7}\)
\(3x-\dfrac{13}{10}=\dfrac{68}{7}\)
\(3x=\dfrac{771}{70}\)
\(x=\dfrac{257}{70}\)
Vậy \(x=\dfrac{257}{70}\)
b)\(\dfrac{75}{100}x-x=-1\dfrac{3}{4}\)
\(-\dfrac{1}{4}x=-\dfrac{7}{4}\)
\(x=7\)
Vậy \(x=7\)
a) Ta có: \(\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{1}{5}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=\dfrac{68}{7}-\dfrac{1}{5}=\dfrac{340}{35}-\dfrac{7}{35}=\dfrac{333}{35}\)
\(\Leftrightarrow x-\dfrac{1}{2}=\dfrac{333}{35}\cdot\dfrac{1}{3}=\dfrac{111}{35}\)
hay \(x=\dfrac{257}{70}\)
Vậy: \(x=\dfrac{257}{70}\)
b) Ta có: \(75\%x-x=-1\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{4}x=-\dfrac{7}{4}\)
hay x=7
Vậy: x=7
75%:2\(\dfrac{1}{5}\)+(0,5)2.(-7)+2,5(7\(\dfrac{2}{3}\)+5\(\dfrac{2}{3}\))
\(75\%:2\dfrac{1}{5}+\left(0,5\right)^2.\left(-7\right)+2,5\left(7\dfrac{2}{3}+5\dfrac{2}{3}\right)\)
\(=\dfrac{3}{4}:\dfrac{11}{5}+\dfrac{1}{4}.\left(-7\right)+2,5\left(12+\dfrac{4}{3}\right)\)
\(=\dfrac{15}{44}-\dfrac{7}{4}+\dfrac{5}{2}.\dfrac{40}{3}\)
\(=-\dfrac{31}{22}+\dfrac{100}{3}\)
\(=\dfrac{2107}{66}\)
Bài 1:
\(a)5\dfrac{1}{20}+\dfrac{8}{9}-\dfrac{15}{25}+\dfrac{75}{-18}\)
\(b)\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(c)\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}.\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
a)\(=\dfrac{211}{180}\)
b)\(=\dfrac{5}{39}\)
c)=\(=-\dfrac{65}{168}\)
e)\(\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
f)\(\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
g)
\(\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
Mn giúp em với ạ
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
a,\(\dfrac{4}{9}\) : \(\dfrac{3}{10}\)
b,\(\dfrac{3}{7}\) x \(\dfrac{5}{7}\)
c,\(\dfrac{4}{9}\) : \(\dfrac{3}{2}\)
D \(\dfrac{3}{9}\) + \(\dfrac{5}{6}\) x \(\dfrac{2}{3}\)
E \(\dfrac{2}{7}\) x \(\dfrac{4}{3}\) + \(\dfrac{2}{6}\)
G \(\dfrac{4}{7}\) : 2 + \(\dfrac{4}{7}\)
a)\(=\dfrac{4}{9}\times\dfrac{10}{3}=\dfrac{40}{27}\)
b)\(\dfrac{3}{7}\times\dfrac{5}{7}=\dfrac{15}{49}\)
c)\(\dfrac{4}{9}:\dfrac{3}{2}=\dfrac{4}{9}\times\dfrac{2}{3}=\dfrac{8}{27}\)
a, =40/27
b, =15/49
c, =8/27
d, =3/9 +5/9
e, = 8/21 +2/6=5/7
g, =2/7+4/7=6/7
tính thuận tiện:
\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\) \(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\) \(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)
\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)
mik sẽ chỉ tick 3 bn xong trước phải chi tiết rõ ràng
a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\)
\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{5}{5}+\dfrac{3}{3}\)
\(=1+1\)
\(=2\)
============
\(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\)
\(=\left(\dfrac{17}{12}+\dfrac{7}{12}\right)+\left(\dfrac{29}{7}-\dfrac{8}{7}\right)\)
\(=\dfrac{24}{12}+\dfrac{21}{7}\)
\(=2+3\)
\(=5\)
====================
\(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)
\(=\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{6}{15}-\dfrac{1}{7}-\dfrac{1}{7}\)
\(=\left(\dfrac{9}{15}+\dfrac{6}{15}\right)+\left(\dfrac{16}{7}-\dfrac{1}{7}-\dfrac{1}{7}\right)\)
\(=\dfrac{15}{15}+\dfrac{14}{7}\)
\(=1+2\)
\(=3\)
===============
\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{5}{5}+\dfrac{3}{3}+\dfrac{8}{4}\)
\(=1+1+2\)
\(=4\)
\(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^3\)
\(4\dfrac{1}{3}.\dfrac{1}{6}-\dfrac{1}{2}.4\dfrac{1}{3}+\left|\dfrac{-13}{3}.\dfrac{7}{15}\right|\)
= 3/4 - 25/6 - 1/8 = -85/24
= 13/18 - 13/6 + 91/45 = 26/45
\(\dfrac{2}{3}.\dfrac{-5}{7}\)
\(\dfrac{-1}{3}.\dfrac{5}{9}\)
\(\dfrac{-2}{5}.\dfrac{10}{4}\)
\(\dfrac{-3}{7}.\dfrac{5}{15}\)
\(\dfrac{-7}{3}.\dfrac{9}{21}\)
a: =-10/21
b: =-5/27
c: =-20/20=-1
d: =-15/105=-1/7
e: =-63/63=-1
\(\dfrac{10}{21}\)
\(\dfrac{5}{27}\)
\(1\)
\(\dfrac{1}{7}\)
\(1\)