K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TA
21 tháng 7 2021
`=(0,125-1/5+1/7)/(3(0,125-1/5+1/7))+(2(1/4+1/6-0,1))/(3(1/4+1/6-0,1))`
`=1/3+2/3=3/3=1`
VQ
3 tháng 9 2023
1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
VQ
0
H
20 tháng 7 2021
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
9 tháng 7 2015
\(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)\(=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{60.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}{60.\left(\frac{3}{4}+\frac{1}{2}-\frac{3}{10}\right)}\)
\(=\frac{1}{3}+\frac{30+20-12}{45+30-18}\)\(=\frac{1}{3}+\frac{38}{57}=1\)
27 tháng 11 2016
\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{2}{4}+\frac{2}{6}-\frac{2}{10}}{\frac{3}{4}+\frac{3}{6}-\frac{3}{10}}\)
= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{2.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}{3.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}\)
= \(\frac{1}{3}+\frac{2}{3}\)
= \(\frac{3}{3}\)
= 1
21 tháng 8 2020
a) \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)=\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\frac{7}{2}\)
\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{1}{3}-1=\frac{1}{8}\cdot\frac{1}{3}-1=\frac{1}{24}-\frac{24}{24}=-\frac{23}{24}\)
b) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)\cdot2\frac{2}{3}\cdot0,25\)
\(=\left(\frac{3}{5}+\frac{83}{200}-\frac{3}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}\)
\(=\left(\frac{3}{5}+\frac{80}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=\left(\frac{3}{5}+\frac{2}{5}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}\)
c) \(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}\cdot8-\frac{33}{20}\cdot\frac{10}{11}=\frac{5}{2}-\frac{3}{2}=1\)
d) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}=\frac{1}{4}:\left(\frac{103}{10}-\frac{98}{10}\right)-\frac{3}{4}\)
\(=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)
Câu cuối tương tự
HN
25 tháng 6 2018
\(1,125-\left(\frac{7}{8}-\frac{2}{3}\right):\frac{4}{3}\)
\(=\frac{1125}{100}-\left(\frac{21}{24}-\frac{16}{24}\right).\frac{3}{4}\)
\(=\frac{45}{4}-\frac{5}{24}.\frac{3}{4}\)
\(=\frac{45}{4}-\frac{15}{32}\)
\(=\frac{360}{32}-\frac{15}{32}\)
\(=\frac{345}{32}\)
25 tháng 6 2018
\(1,125-\left(\frac{7}{8}-\frac{2}{3}\right)\div\frac{4}{3}\)
\(=\frac{9}{8}-\left(\frac{7}{8}-\frac{2}{3}\right)\div\frac{4}{3}\)
\(=\frac{9}{8}-\frac{5}{24}\div\frac{4}{3}\)
\(=\frac{9}{8}-\frac{5}{32}\)
\(=\frac{31}{32}\).
~ CHÚC BẠN HOK TỐT ~
\(\dfrac{-3}{10}-0,125+\dfrac{-7}{10}+1,125\)
\(=(\dfrac{-3}{10}+\dfrac{-7}{10})+(-0,125+1,125)\)
\(=\dfrac{-10}{10}+1\)
\(=-1+1=0\)
`(-3)/10 - 0,125 + (-7)/10 + 1,125`
`=(-3)/10- 1/8 -7/10 + 9/8`
`=(3/10 - 7/10) + (-1/8 + 9/8)`
`=-10/10 + 8/8`
` = -1 + 1`
`=0`