38 + −34 
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15 tháng 8 2019

\(a,\left(38+-34+712\right):56+12\)

\(=\left(4+712\right):56+12\)

\(=716:56+12\)

\(=\frac{179}{14}+12\)

\(=\frac{347}{14}\)

\(b,12+34-\left(34-45\right)\)

\(=12+34--11\)

\(=46--11\)

\(=57\)

\(c,6512:234+1114\cdot\left(13-15\right)\)

\(=6512:234+1114\cdot-2\)

\(=\frac{3256}{117}+-2228\)

\(=\frac{-257420}{117}\)

\(d,\left(78-34\right)\cdot113-27\cdot\left(3,5\right)^2\)

\(=44\cdot113-27\cdot\frac{49}{4}\)

\(=4972-\frac{1323}{4}\)

\(=\frac{18565}{4}\)

\(e,\left(35+0,415-3200\right)\cdot223\cdot0,25\)

\(=-3164,585\cdot223\cdot0,25\)

\(=-176425,6138\)

\(f,516:0,125-\left(214-0,6\right)\cdot1011\)

\(=516:0,125-213,4\cdot1011\)

\(=4128-215747,4\)

\(=-211619,4\)

\(g,0,25:\left(10,3-9,8\right)-34\)

\(=\frac{1}{4}:\frac{1}{2}-34\)

\(=\frac{1}{2}-34\)

\(=\frac{-67}{2}\)

\(h,11315\cdot0,75-\left(1120+25\%\right):73\)

\(=11315\cdot0,75-\frac{4481}{4}:73\)

\(=\frac{33945}{4}-\frac{4481}{292}\)

\(=\frac{618376}{73}\)

\(k,23+27-114-1-37+328\)

\(=50-114-1-37+328\)

\(=-64-1-37+328\)

\(=-65-37+328\)

\(=-102+328\)

\(=226\)

MK CHUC BAN HOC TOT :))

7 tháng 3

a)\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)

=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{14}{24}\right)\times\frac65+\frac12\)

=\(\frac{5}{24}\times\frac65+\frac12\)

=\(\frac14+\frac12\)

=\(\frac14+\frac24\)

=\(\frac34\)

7 tháng 3

b)\(\frac12+\frac34-\left(\frac34-\frac45\right)\)

=\(\frac12+\frac34-\frac34+\frac45\)

=\(\left(\frac12+\frac45\right)+\left(\frac34-\frac34\right)\)

=\(\left(\frac{5}{10}+\frac{8}{10}\right)+0\)

=\(\frac{13}{10}\)

4 tháng 3 2017

a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)

= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)

= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)

= \(\dfrac{7}{13}\times\dfrac{7}{12}\)

= \(\dfrac{49}{156}\)

b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)

= 0

24 tháng 3 2018

a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)

b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)

\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)

c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)

d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)

e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)

\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)

f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)

g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)

h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)

i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)

22 tháng 10 2019

a) \(47-\left[\left(45\cdot2^4-5^2\cdot12\right):14\right]\)

\(=47-\left[\left(45\cdot16-25\cdot12\right):14\right]\)

\(=47-\left[\left(720-300\right):14\right]\)

\(=47-\left[420:14\right]\)

\(=47-30=17\)

b) \(50-\left[\left(20-2^3\right):2+34\right]\)

\(=50-\left[\left(20-8\right):2+34\right]\)

\(=50-\left[12:2+34\right]\)

\(=50-\left[6+34\right]\)

\(=50-40=10\)

c) \(10^2-\left[60:\left(5^6:5^4-3\cdot5\right)\right]\)

\(=100-\left[60:\left(5^{6-4}-15\right)\right]\)

\(=100-\left[60:\left(5^2-15\right)\right]\)

\(=100-\left[60:\left(25-15\right)\right]\)

\(=100-\left[60:10\right]\)

\(=100-6=94\)

d) \(50-\left[\left(50-2^3\cdot5\right):2+3\right]\)

\(=50-\left[\left(50-8\cdot5\right):2+3\right]\)

\(=50-\left[\left(50-40\right):2+3\right]\)

\(=50-\left[10:2+3\right]\)

\(=50-\left[5+3\right]\)

\(=50-8=42\)

e) \(10-\left[\left(8^2-48\right)\cdot5+\left(2^3\cdot10+8\right)\right]:28\)

\(=10-\left[\left(64-48\right)\cdot5+\left(8\cdot10+8\right)\right]:28\)

\(=10-\left[16\cdot5+\left(80+8\right)\right]:28\)

\(=10-\left[80+88\right]:28\)

\(=10-168:28\)

\(=10-6=4\)

f) \(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)

\(=8697-\left[3^{7-5}+2\cdot10\right]\)

\(=8697-\left[3^2+20\right]\)

\(=8697-\left[9+20\right]\)

\(=8697-29=8668\)

CHÚC BẠN HỌC TỐT!!!!!!!!!!!

2 tháng 7 2019

\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)

\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)

\(=\frac{5}{2}-\frac{33}{20}\cdot\frac{10}{11}\)

\(=\frac{5}{2}-\frac{3}{2}\)

\(=1\)

\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}:\frac{125}{1000}-\left(\frac{9}{4}-\frac{6}{10}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}.\frac{8}{1}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)

\(\Rightarrow\frac{5}{2}-\frac{3.1}{2.1}\)

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

a: \(=\dfrac{1}{4}\cdot\dfrac{12}{5}\cdot\dfrac{100}{7}\cdot\dfrac{49}{100}\)

\(=\dfrac{1}{4}\cdot\dfrac{12}{5}\cdot\dfrac{49}{7}=\dfrac{3}{5}\cdot7=\dfrac{21}{5}\)

b: \(=\dfrac{3}{8}+\dfrac{1}{8}\cdot\dfrac{3}{4}-\dfrac{5}{4}\)

\(=\dfrac{12}{32}+\dfrac{3}{32}-\dfrac{40}{32}=\dfrac{-25}{32}\)

c: \(=\dfrac{4}{9}\left(\dfrac{-13}{27}-\dfrac{14}{27}\right)-\dfrac{5}{9}=\dfrac{-4}{9}-\dfrac{5}{9}=-1\)

d: \(=\dfrac{2}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{91\cdot95}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{91}-\dfrac{1}{95}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{92}{285}=\dfrac{46}{285}\)

30 tháng 4 2018

a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)

\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)

\(=0-\frac{19}{26}\)

\(=-\frac{19}{26}\)

30 tháng 4 2018

c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=-1\)