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1) 2300 : 50 = ( 2300 . 2 ) : ( 50 . 2 ) = 4600 : 100 = 46
2) 112 : 2 = ( 112 . 5 ) : ( 2 . 5 ) = 560 : 10 = 56
3) 2300 : 4 = ( 2300 . 25 ) : ( 4 . 25 ) = 57500 : 100 = 575
4) 3500 : 25 = ( 3500 . 4 ) : ( 4 . 25 ) = 14000 : 100 = 140
5) 1200 : 75 = ( 1200 . 4 ) : ( 4 . 75 ) = 4800 : 300 = 6
6) 2100 : 75 = ( 2100 . 4 ) : ( 4 . 75 ) = 8400 : 300 = 28
7) 320 : 8 = ( 320 . 5 ) : ( 8 . 5 ) = 1600 : 40 = 40
1) 2300 : 50 = (2300.2) : (50.2) = 4600 : 100 = 46
2) 112: 2 = ( 112.5) : (2.5) = 560 : 10 = 56
3) 2300 : 4 = ( 2300.25) : (4.25) = 57500 : 100 = 575
4)3500 : 25 = (3500.4) : (25.4) = 14000 : 100 = 140
5)1200 : 75 = ( 1200 : 3) : (75:3) = 400 : 25 = (400.4) : (25.4) = 1600 : 100 = 16
...
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Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$
$=2.3+2^3.3+...+2^{99}.3$
$=3(2+2^3+...+2^{99})\vdots 3$
Ta có đpcm.
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1) [ 179 - (x - 13) ] : 11 = 14
179 - (x - 13) = 14 . 11
179 - (x - 13) = 154
x - 13 = 179 - 154
x - 13 = 25
x = 25 + 13
x = 38
2) 2100 : 50 = 42
1400 : 25 = 56
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\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(M< \dfrac{1}{4}+\left(\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}+M_1\)
\(M_1=\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)...+\left(\dfrac{1}{48}-\dfrac{1}{49}\right)+\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(M_1=\dfrac{1}{2}+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...+\left(-\dfrac{1}{49}+\dfrac{1}{49}\right)-\dfrac{1}{50}=\dfrac{1}{2}-\dfrac{1}{50}\)
\(M< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{3}{4}-\dfrac{1}{50}< \dfrac{3}{4}=>dpcm\)
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