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a: \(20,11\cdot7,5+20,11+20,11+20,11\cdot0,5\)
\(=20,11\left(7,5+1+1+0,5\right)\)
\(=20,11\cdot10=201,1\)
b: \(2,5\cdot12,85\cdot4\)
\(=12,85\cdot\left(2,5\cdot4\right)\)
\(=12,85\cdot10=128,5\)
a: 20,11⋅7,5+20,11+20,11+20,11⋅0,520,11⋅7,5+20,11+20,11+20,11⋅0,5
=20,11(7,5+1+1+0,5)=20,11(7,5+1+1+0,5)
=20,11⋅10=201,1=20,11⋅10=201,1
b: 2,5⋅12,85⋅42,5⋅12,85⋅4
=12,85⋅(2,5⋅4)=12,85⋅(2,5⋅4)
=12,85⋅10=128,5
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số số hạng là :`(26 - 2) : 2 + 1 = 13(số hạng)`
`13 : 2 = 12 dư 1 `
`=>A = (2 - 4) + (6 - 8) + (10 - 12) + ... + (22 - 24) + 26`
`=>A = (-2) + (-2) + (-2) + ... + (-2) + 26`
`=>A = (-2) xx 6 + 26`
`=> A = -12 + 26`
`=> A = 14`
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Lời giải:
$A=(99-97)+(95-93)+(91-89)+...+(7-5)+(8-1)$
$=\underbrace{2+2+2+...+2}_{24}+7$
$=2\times 24+7$
$=55$
Ta có: 99–97 +95–93+...+7–5+3−1
=(99−97)+(95−93)+...+(7−5)+(3−1)
=2+2+...+2+2 (có 25 số 2)
=2.25
=50
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`A=1/[1xx2xx3]+1/[2xx3xx4]+1/[3xx4xx5]+....+1/[98xx99xx100]`
`A=1/2xx(2/[1xx2xx3]+2/[2xx3xx4]+2/[3xx4xx5]+....+2/[98xx99xx100])`
`A=1/2xx(1/[1xx2]-1/[2xx3]+1/[2xx3]-1/[3xx4]+1/[3xx4]-1/[4xx5]+....+1/[98xx99]-1/[99xx100])`
`A=1/2xx(1/[1xx2]-1/[99xx100])`
`A=1/2xx(1/2-1/9900)`
`A=1/2xx(4950/9900-1/9900)`
`A=1/2xx4949/9900`
`A=4949/19800`
\(A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}\)
\(A=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right):2\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{12}-\dfrac{1}{20}+...+\dfrac{1}{9702}-\dfrac{1}{990}\right):2\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{990}\right):2\)
\(A=\dfrac{4949}{9900}:2\)
\(A=\dfrac{4949}{19800}\)
25.(187+18+1382)
= 25.(187+1400)
=25.1400+187
=3500 +187
= 35187