Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
a, 387 . 2694 + 7305 . 387 + 387
= 387 . ( 2694 + 7305 ) +387
= 387 . 9999 + 387
= 3869613 + 387
= 3870000
hai câu còn lại tự làm
Ta có : \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}\right).\frac{2005}{1890}+115\)
\(=\left(\frac{3}{5}-\frac{3}{5}\right).\frac{2005}{1890}+115=0+115=115\)
= ( \(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)+ \(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)) x \(\frac{2005}{1890}\)+ 115
= ( \(\frac{3(\frac{1}{2}+\frac{1}{3}-\frac{1}{4})}{5(\frac{1}{1}+\frac{1}{3}-\frac{1}{4})}\)+ \(\frac{3(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}{-5(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}\)) x \(\frac{2005}{1890}\)+ 115
=( \(\frac{3}{5}\)+\(\frac{3}{-5}\)) x \(\frac{2005}{1890}\)+115 = 0 +115 = 115
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)
a,[15:(-3)+40:(-8)]-3(16:8)
= ( - 5 - 5 ) - 3. 2
= - 10 - 6
= - 16
b, 3 ( 25 : 5 - 14 : 2 ) - 5 ( 6 : 2 )
= 3 . (5 - 7 ) - 5 . 3
= 3 . - 2 - 5 . 3
= - 21
c, [ - 15 : ( - 3 ) ] - 3 [ 2. ( 5 - 9 : 3 ) ]
= 5 - 3 . 4
= - 7
d, - 2 ( - 3 + 8 : 2 - 7 ) - 5 . 4
= - 2 . - 6 - 5 . 4
= 12 - 20
= - 8
e, 7 . [ - 6 : 2 - 15 : ( - 3 ) - 3 ]
= 7 . ( - 3 + 5 - 3)
= - 7
f, 17 . ( - 84 ) + 17 . ( - 16)
= - 1428 - 272
= - 1700
a,[15:(-3)+40:(-8)]-3(16:8)
= ( - 5 - 5 ) - 3. 2
= - 10 - 6
= - 16
b, 3 ( 25 : 5 - 14 : 2 ) - 5 ( 6 : 2 )
= 3 . (5 - 7 ) - 5 . 3
= 3 . - 2 - 5 . 3
= - 21
c, [ - 15 : ( - 3 ) ] - 3 [ 2. ( 5 - 9 : 3 ) ]
= 5 - 3 . 4
= - 7
d, - 2 ( - 3 + 8 : 2 - 7 ) - 5 . 4
= - 2 . - 6 - 5 . 4
= 12 - 20
= - 8
e, 7 . [ - 6 : 2 - 15 : ( - 3 ) - 3 ]
= 7 . ( - 3 + 5 - 3)
= - 7
f, 17 . ( - 84 ) + 17 . ( - 16)
= - 1428 - 272
= - 1700
1) \(35.18-5.7.28\)
\(=35.18-35.28\)
\(=35.\left(18-28\right)\)
\(=35.\left(-10\right)\)
\(=-350\)
2) \(45-5.\left(12+9\right)\)
\(=45-60-45\)
\(=\left(45-45\right)-60\)
\(=0-60\)
\(=-60\)
3) \(24.\left(16-5\right)-16.\left(24-5\right)\)
\(=\left(384-120\right)-\left(384-120\right)\)
\(=264-264\)
\(=0\)
4) \(13.\left(23+22\right)-3.\left(17+28\right)\)
\(=13.45-3.45\)
\(=\left(13-3\right).45\)
\(=10.45\)
\(=450\)
tk ủng hộ nha!!!!!!!!1
1/ 35 . 18 - 5 . 7 . 28
= 35 . 18 - 35 . 28
= 35 . (18 - 28)
= 35 . (-10)
= -350
2/ 45 - 5 . (12 + 9)
= 5 . 9 - 5 . 12 + 5 . 9
= 5 . (9 - 12 + 9)
= 5 . 6
= 30
3/ 24 . (16 - 5) - 16 . (24 - 5)
= 24 . 16 - 24 . 5 - 16 . 24 - 16 . 5
= (24 . 16 - 16 . 24) - (24 . 5 - 16 . 5)
= 0 - [5 . (24 - 16)]
= 0 - [5 . 8]
= 0 - 40
= -40
4/ 13 . (23 + 22) - 3 . (17 + 28)
= 13 . 45 - 3 . 45
= 45 . (13 - 3)
= 45 . 10
= 450
a)\(\frac{-5}{18}+\frac{32}{45}-\frac{9}{10}\)
\(=\frac{-25}{90}+\frac{64}{90}-\frac{81}{90}\)
\(=-\frac{42}{90}=-\frac{7}{15}\)
b)\(\frac{3}{4}-\left(-\frac{5}{8}\right)+\frac{9}{4}-\frac{3}{16}\)
\(=\frac{3}{4}+\frac{5}{8}+\frac{9}{4}-\frac{3}{16}\)
\(=\frac{12}{16}+\frac{10}{16}+\frac{36}{16}-\frac{3}{16}\)
\(=\frac{58}{16}=\frac{29}{8}\)
c) \(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}=\frac{11}{125}\)
d) \(\frac{-2}{5}-\frac{3}{10}-\left(-\frac{1}{2}\right)\)
\(=-\frac{2}{5}-\frac{3}{10}+\frac{1}{2}\)
\(=-\frac{4}{10}-\frac{3}{10}+\frac{5}{10}\)
\(=-\frac{2}{10}=-\frac{1}{5}\)
a: \(=\dfrac{15}{4}:\dfrac{2-7}{16}-\dfrac{5}{9}\cdot\left(\dfrac{3}{5}+\dfrac{4}{5}\right)\)
\(=\dfrac{15}{4}\cdot\dfrac{16}{-5}-\dfrac{5}{9}\cdot\dfrac{7}{5}\)
\(=\dfrac{-240}{20}-\dfrac{7}{9}=-12-\dfrac{7}{9}=\dfrac{-115}{9}\)
c: \(=\dfrac{1}{5}-\dfrac{7}{2}-\dfrac{3}{2}+\dfrac{5}{4}\)
\(=\dfrac{4+25}{20}-5=\dfrac{29}{20}-\dfrac{100}{20}=\dfrac{-71}{20}\)
d: \(=\dfrac{12}{17}\left(1-\dfrac{1}{15}-\dfrac{4}{5}+1\right)=\dfrac{12}{17}\cdot\dfrac{17}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
e: \(=\dfrac{2}{15}\cdot\dfrac{9}{8}-\dfrac{7}{4}\cdot\dfrac{9}{8}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\left(\dfrac{2}{15}-\dfrac{7}{4}\right)+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{8-105}{60}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{-97}{60}+\dfrac{25}{36}=\dfrac{-1619}{1440}\)