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1) 40 + 15 + (-10) + (-15) 2) -13 + (-750) + (-17) + 750 3) (35 - 17) + (17 + 120 - 35)
= 40 + 15 - 10 - 15 = -13 - 750 - 17 + 750 = 35 - 17 + 17 + 120 - 35
= (40 - 10) + (15 - 15) = (-13 - 17) + (-750 + 750) = (35 - 35) + (-17 + 17) + 120
= 30 = -30 = 120
4) (55 + 45 + 15) - (15 - 55 + 45) 5) -(12 + 21 - 23) - (23 - 21 + 10) 6) (2020 - 79 + 15) - (-79 + 15)
= 55 + 45 + 15 - 15 + 55 - 45 = -12 -21 + 23 - 23 + 21 - 10 = 2020 - 79 + 15 + 79 - 15
= (45 - 45) + (15 - 15) + (55 + 55) = (-12 - 10) + (-21 + 21) + (23 - 23) = 2020 + (-79 + 79) + (15 - 15)
= 110 = -22 = 2020
7) -(515 - 80 + 91) - (2010 + 80 - 91) 8) 25 - (-17) + 24 - 12 9) 235 - (34 + 135) - 100
= -515 + 80 - 91 - 2010 - 80 + 91 = 25 + 17 + 24 - 12 = 235 - 34 - 135 - 100
= (-515 -2010) + (80 - 80) + (-91 + 91) = 54 = -34
= -2525
10) (13 + 49) - (13 - 135 + 49)
= 13 + 49 - 13 + 135 - 49
= (13 - 13) + (49 - 49) +135
= 135
Dấu " . " k p dấu " , "
a) \(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+\frac{91}{7\cdot10}+...+\frac{91}{88\cdot99}\)
* Bài đúng k z ???
b) \(5\frac{1}{7}\left(3\frac{2}{3}+4\frac{1}{7}\right)=\frac{36}{7}\cdot\left(\frac{11}{3}+\frac{29}{7}\right)\)
\(=\frac{36}{7}\cdot\frac{164}{21}=\frac{1968}{49}\)
\(VT=91\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{88\cdot91}\right)\)
\(=\dfrac{91}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{88\cdot91}\right)\)
\(=\dfrac{91}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{88}-\dfrac{1}{91}\right)\)
\(=\dfrac{91}{3}\cdot\dfrac{90}{91}=30\)
a)36:32+23.22
=\(3^{6-2}+2^{3+2}=3^4+\)\(2^5\)
=81 + 32 = 113
b) \(\left(39.42-37.42\right):42=42\left(39-37\right):42\)=2
c) \(5.4^2-18:3^2=5.16-18:9=80-2=78\)
d) \(80-\left[130-\left(12-4\right)^2\right]=80-\left(130-8^2\right)=80-\left(130-64\right)\)
= 80 - 66=14
80/1.6+80/6.11+80/11.16+...+80/251.256
=16.(5/1.6+5/6.11+5/11.16+...+5/251.256
=16.(1-1/6+1/6-1/11+1/11-1/16+...+1/251-1/256)
=16.(1-1/256)
=16.255/256
=255/16
a/=(74-(-1937)1)
=74-(-1937)
=2011
b/=4/7+5/6:5-3/8*(-4)
=4/7+1/6-(-3/2)
=31/42-(-3/2)
=47/21
minh chi biet bay nhieu
Ta có :
\(\frac{7}{12}\)= \(\frac{4}{12}\)+ \(\frac{3}{12}\)= \(\frac{1}{3}\)+ \(\frac{1}{4}\)= \(\frac{20}{60}\)+ \(\frac{20}{80}\)
\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ .... + \(\frac{1}{79}\)+ \(\frac{1}{80}\)= (\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ ....+\(\frac{1}{60}\)) + ( \(\frac{1}{61}\)+ \(\frac{1}{62}\)+...+\(\frac{1}{79}\)+\(\frac{1}{80}\))
Do \(\frac{1}{41}\)>\(\frac{1}{42}\)>....>\(\frac{1}{60}\)
=> ( \(\frac{1}{41}\)+ \(\frac{1}{42}\)+...+\(\frac{1}{60}\)) > \(\frac{1}{60}\)+...+\(\frac{1}{60}\)= \(\frac{20}{60}\)
Vậy : \(\frac{1}{61}\)> \(\frac{1}{62}\)>....>\(\frac{1}{79}\)>\(\frac{1}{80}\)
=> ( \(\frac{1}{61}\)+\(\frac{1}{62}\)+...+\(\frac{1}{79}\)+ \(\frac{1}{80}\)) > \(\frac{1}{80}\)+...+ \(\frac{1}{80}\)= \(\frac{20}{80}\)
Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+\(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)
Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+ \(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)= \(\frac{7}{12}\)
=> ĐPCM
Tham Khảo:
-(515-80+91)-(2003+80-91)
=-515+80-91-2003-80+91
(-515)-2003+(80-80)+(91-91)
=2518