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a: \(232-\left(581+132-331\right)\)
\(=232-581-132+331\)
\(=\left(232-132\right)+\left(-581+331\right)\)
\(=100-250\)
=-150
b: \(\left[12+\left(-57\right)\right]-\left[-57-\left(-12\right)\right]\)
\(=12+\left(-57\right)+57+\left(-12\right)\)
\(=\left(12-12\right)+\left(57-57\right)\)
=0+0
=0
a) \(A=27\cdot36+73\cdot99+27\cdot14-49\cdot73\)
\(A=27\cdot\left(36+14\right)+73\cdot\left(99-49\right)\)
\(A=27\cdot50+73\cdot50\)
\(A=50\cdot\left(27+73\right)\)
\(A=50\cdot100\)
\(A=5000\)
b) \(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)
\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B-\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)
\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)
\(B=2^3\cdot5^3\)
\(B=10^3\)
\(B=1000\)
a: =382-282+531-331
=100+200=300
b: =(7-8)+(9-10)+...+(2009-2010)
=(-1)+(-1)+....+(-1)
=-1002
c: =-(1+2+3+...+2009+2010)
=-2010*2011/2=-2021055
\(\dfrac{1}{2}\cdot1\cdot\dfrac{1}{3}\cdot10\cdot\dfrac{7}{35}\cdot\dfrac{3}{4}\)
\(=\dfrac{1}{2}\cdot10\cdot\dfrac{1}{5}\cdot1\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\)
\(=\dfrac{10}{10}\cdot\dfrac{1}{4}=\dfrac{1}{4}\)
Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
`28 xx 7,32 - 7,32 : 0,125`
`= 28 xx 7,32 - 7,32 xx8`
`= 7,32 xx(28-8)`
`= 7,32 xx20`
`=146,4`