a x 101 - a + 72 x 99 + 72 = 21500
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a) \(101x-x+72.99+72=21050\)
\(100x+72.\left(99+1\right)=21050\)
\(100x+72.100=21050\)
\(\left(x+72\right).100=21050\)
\(x+72=210,5\)
\(x=138,5\)
b) \(x:4+x:0,125=9,9\)
\(x:\left(4+0,125\right)=9,9\)
\(x:4,125=9,9\)
\(x=40,8375\)
c) \(1,5x-5,7=5,3\)
\(1,5x=11\)
\(x=\frac{22}{3}\)
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Xx101-X+72*99+72=21050
Xx100+72*99+72=21050
Xx100+7200=21050
Xx100=21050-7200
Xx100=13850
X=13850:100
X=138,5
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Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
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a) A = 2 + 4 + 6 + 8 + ... + 1000
Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )
= (1000 + 2) . 500 : 2 = 250500
c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
a x 101 - a + 72 x 99 + 72 = 21500
a x 101 - a + 72 x 99 + 72 = a x 101 - a x 1 + 72 x 99 + 72 x 1 = a x (101 - 1) + 72 x (99 + 1) = a x 100 + 72 x 100 = (a + 72) x 100 = 21500.
a + 72 = 21500 : 100 = 215.
a + 72 = 215=> a = 215-72
a=143
a x 101 - a + 72 x 99 + 72 = 21500
a x ( 101 - 1 ) + 72 x ( 99+ 1 ) = 21500
a x 100 + 72 x 100 = 21500
a x 100 + 7200 = 21500
a x 100 = 21500 - 7200
a x 100 =14300
a = 143 00 : 100
a = 143