Tính 1 cách hợp lí:
a, 4.{3^2.[(5^2+2^3):11]-26}+2002
tính một cách hợp lí:
a) \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
b) \(\left(\dfrac{-1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(\dfrac{-15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
c) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
giải chi tiết giúp mình nha
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: =−2590+6490−8190=−2590+6490−8190
(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−53+4229=−53+4229
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
Tính một cách hợp lí:
a) 2 834 + 275 - 2 833 - 265;
b) (11 + 12 + 13) - (1 + 2 + 3).
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) 2834 + 275 - 2833 - 265= 2833+1+275-2833-265=0+1+275-265=11
b) (11 + 12 + 13) - (1 + 2 + 3)=11 + 12 + 13-1-2-3=10+10+10=30
a) 2 834 + 275 - 2 833 - 265
=(275 - 265) + (2 834 - 2 833)
=10 + 1 = 11
b) (11 + 12 +13 ) - (1 + 2 + 3)
= 11 + 12 +13 - 1 - 2 - 3
= (11 - 1) + (12 - 2) +(13 - 3)
=10 +10 +10 =30
Tính theo cách hợp lí:
a) (-1)+2+(-3)+4+....+(-2009)+2010
b) 1+(-2)+(-3)+4+5+(-6)+(-7)+8+....+.....+2005+(-2006)+(-2007)+2008.
giúp với ạ!!
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
a) Ta có: \(\left(-1\right)+2+\left(-3\right)+4+...+\left(-2009\right)+2010\)
\(=1+1+...+1\)
=1005
b) Ta có: \(1+\left(-2\right)+\left(-3\right)+4+5+\left(-6\right)+\left(-7\right)+8+...+2005+\left(-2006\right)+\left(-2007\right)+2008\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2005-2006-2007+2008\right)\)
=0
Bài 1.( 2 điểm)Tính bằng cách hợp lí:
a) \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
b) \(\left(\dfrac{18}{23}+\dfrac{7}{12}\right)+\left(\dfrac{-13}{19}-\dfrac{3}{4}\right)+\left(\dfrac{-6}{19}+\dfrac{5}{23}\right)\)
c) \(\dfrac{4}{3}+\dfrac{-5}{6}+\dfrac{-1}{4}\)
d) \(\dfrac{5}{6}-\dfrac{7}{5}+\dfrac{17}{30}\)
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
Tính một cách hợp lí:
a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\);
b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6};\)
c) \(\dfrac{-5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}.\)
\(a.\)
\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)
\(b.\)
\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)
\(c.\)
\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)
Tính một cách hợp lí:
a)\(\frac{{ - 3}}{{10}} - 0,125 + \frac{{ - 7}}{{10}} + 1,125\); b)\(\frac{{ - 8}}{3}.\frac{2}{{11}} - \frac{8}{3}:\frac{{11}}{9}\);
a)
\(\begin{array}{l}\frac{{ - 3}}{{10}} - 0,125 + \frac{{ - 7}}{{10}} + 1,125 \\= \left( {\frac{{ - 3}}{{10}} + \frac{{ - 7}}{{10}}} \right) + \left( {1,125 - 0,125} \right)\\ = - 1 + 1 \\= 0\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 8}}{3}.\frac{2}{{11}} - \frac{8}{3}:\frac{{11}}{9} \\= \frac{8}{3}.\frac{{ - 2}}{{11}} - \frac{8}{3}.\frac{9}{{11}}\\ = \frac{8}{3}.\left( {\frac{{ - 2}}{{11}} - \frac{9}{{11}}} \right)\\ =\frac{{ - 8}}{3}.\frac{-11}{11}\\= \frac{8}{3}.\left( { - 1} \right) \\= \frac{{ - 8}}{3}\end{array}\)
câu A 2^4-50:25+13.7
Câu B 4.{ 3^2.[(5^2+2^3):11^1]-26}+2002
Câu a.
\(2^4-50:25+13.7=16-2+91=105\)
Câu b.
\(4.\left\{3^2.\left[\left(5^2+2^3\right):11^1\right]-26\right\}+2002\)
\(=4.\left\{9.\left[\left(25+8\right):11\right]-26\right\}+2002\)
\(=4.\left[9.\left(33:11\right)-26\right]+2002\)
\(=4.\left(9.3-26\right)+2002=4.\left(27-26\right)+2002=4.1+2002=2006\)
tính
4 . { 32 - [ ( 52 + 23 ) : 111 ] - 26 } + 2002
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
gấp nha mn, mk thik cho
Bài 3.29
Tính một cách hợp lí:
a) 2 834 + 275 - 2 833 - 265
b) (11 + 12 + 13) - (1 + 2 + 3)
ai cùng câu hỏi của mk thì thik câu hỏi cho mk biết nha
a) 2 834 + 275 - 2 833 - 265= ( 2834-2833)+ (275-265 ) = 1 + 10 = 11
b) (11 + 12 + 13) - (1 + 2 + 3)= 11 + 12 + 13 -1-2-3 = ( 11-1)+(12-2)+(13-3) = 10 + 10 + 10 = 30