Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
Tính một cách hợp lí:
a) 2 834 + 275 - 2 833 - 265;
= (2 834 – 2 833) + (275 – 265)
= 1 + 10
= 11
b) (11 + 12 + 13) - (1 + 2 + 3).
= (11 – 1) + (12 – 2) + (13 – 3)
= 10 + 10 + 10
= 30
a) (-1) + 2 + (-3) + 4 + .... + (-2009) + 2010
= (-1 + 2) + (-3 + 4) + ..... + (-2009 + 2010)
= -1 + (-1) + (-1) + .... + (-1)
= -1 . 1005 = -1005
b) 1 + (-2) + (-3) + 4 + 5 + (-6) + (-7) + 8 + ... + 2005 + (-2006) + (-2007) + 2008
= [1 + (-2) + (-3) + 4] + [5 + (-6) + (-7) + 8 ] + ..... + [2005 + (-2006) + (-2007) + 2008]
= 0 + 0 + ...... + 0 = 0
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
\(a.\)
\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)
\(b.\)
\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)
\(c.\)
\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)
Câu a.
\(2^4-50:25+13.7=16-2+91=105\)
Câu b.
\(4.\left\{3^2.\left[\left(5^2+2^3\right):11^1\right]-26\right\}+2002\)
\(=4.\left\{9.\left[\left(25+8\right):11\right]-26\right\}+2002\)
\(=4.\left[9.\left(33:11\right)-26\right]+2002\)
\(=4.\left(9.3-26\right)+2002=4.\left(27-26\right)+2002=4.1+2002=2006\)
Ko viết lại đề bài nha
= 4 . { 9 - [ ( 25 + 8 ) : 11 ] - 26 } + 2002
= 4. { 9 - [ ( 33 : 11 ] - 26 } + 2002
= 4 . { 9 - 3 - 26 } + 2002
= 4. { 6 - 26 } + 2002
= 4. - 20 + 2002
= -80 + 2002
= 1922
1:
a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)
b:
Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\)
\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)
Trả lời :
\(a,4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2002\)
\(=4.\left\{9.\left[\left(25+8\right):11\right]-26\right\}+2002\)
\(=4.\left\{9.\left[33:11\right]-26\right\}+2002\)
\(=4.\left\{9.3-26\right\}+2002\)
\(=4.1+2002\)
\(=2006\)
Hok_tốt
#Thiên_Hy
\(4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2002\)
\(=4.\left\{9.\left[\left(25+8\right):11\right]-26\right\}+2002\)
\(=4.\left\{9.\left[33:11\right]-26\right\}+2002\)
\(=4.\left\{9.3-26\right\}+2002\)
\(=4.\left\{27-26\right\}+2002\)
\(=4.1+2002\)
\(=4+2002\)
\(=2006\)
~ Hok tốt ~