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152 + (-173) - (-18) - 127
= 152 + (-173) + 18 - 127
= (152 + 18) + (-173) - 127
= 170 + (-300)
= -130
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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
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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
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\(\frac{1978.1979+1980.21+1958}{1980.1979-1978.1979}\)
= \(\frac{1978.1979+1979.21+21+1958}{1979.\left(1980-1978\right)}\)
= \(\frac{1978.1979+1979.21+1979}{1979.2}\)
= \(\frac{1979.\left(1978+21+1\right)}{1979.2}\)
= \(\frac{1978+21+1}{2}\)
= \(\frac{2000}{2}\)= 1000
=1979x1978+1979x21+21+1958/1979x(1980-1978)
=1979x1979+1979/1979x2
=1979x1980/1979x2
=1979x2x990/1979x2
=990 chúc bạn học tốt nha
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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
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\(A=\frac{27.45+55.27}{2+4+6+...+18}\);
\(A=\frac{27.\left(45+55\right)}{\frac{\left(18+2\right).9}{2}}\)
\(A=\frac{27.100}{90}\)
\(A=\frac{2700}{90}\)
\(A=30\)
\(\Rightarrow A=\frac{27x\left(45+55\right)}{2+4+6+....+18}\)
\(\Rightarrow A=\frac{27x100}{2+4+6+..+18}\)
\(\Rightarrow A=\frac{2700x}{2+4+6+...+18}\)
Dặt M = 2+4+6+...+18
Số số hạng M là : ( 18 - 2 ):2 +1 =9 ( số hạng )
Tổng M = \(\frac{\left(18+2\right).9}{2}=90\)
Thay M vào ta có : \(A=\frac{2700x}{90}=30x\)
Chúc bạn hok tốt !