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\(46\cdot99\)
\(=46\cdot\left(100-1\right)\)
\(=46\cdot100-46\)
\(=4600-46\)
\(=4554\)
___________
\(34\cdot11\)
\(=34\cdot\left(10+1\right)\)
\(=34\cdot10+34\)
\(=340+34\)
\(=374\)
___________
\(25\cdot12\)
\(=5\cdot5\cdot12\)
\(=5\cdot60\)
\(=300\)
__________
\(15\cdot4\)
\(=15\cdot2\cdot2\)
\(=30\cdot2\)
\(=60\)
____________
\(45\cdot6\)
\(=45\cdot2\cdot3\)
\(=90\cdot3\)
\(=2700\)
________
\(13\cdot99\)
\(=13\cdot\left(100-1\right)\)
\(=13\cdot100-13\)
\(=1300-13\)
\(=1287\)
______________
\(16\cdot19\)
\(=16\cdot\left(20-1\right)\)
\(=16\cdot20-16\)
\(=320-16\)
\(=304\)
_____________
\(35\cdot98\)
\(=35\cdot\left(100-2\right)\)
\(=35\cdot100-35\cdot2\)
\(=3500-70\)
\(=3430\)
____________
\(125\cdot16\)
\(=125\cdot8\cdot2\)
\(=1000\cdot2\)
\(=2000\)
___________
\(47\cdot101\)
\(=47\cdot\left(100+1\right)\)
\(=47\cdot100+47\)
\(=4700+47\)
\(=4747\)
46 x 99 =4554 , 34 x 11 = 374 , 25 x 12 =- 300 , 15 x 4 = 60 , 45 x 6 = 270 , 13 x 99 = 1287 , 16 x 19 = 304
35 x 98 = 3430 , 125 x 16 = 2000 , 47 x 101 = 4747
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a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
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47. (23 + 50)- 23. ( 47+50 )
= 47. 23+ 47. 50- 23. 47 - 23. 50
=47. 50 - 23. 50
= 50 .( 47 -23)
=50. 24 = 1200
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a, \(x\) + 99: 3 = 55
\(x\) + 33 = 55
\(x\) = 55 - 33
\(x\) = 22
b, (\(x\) - 25):15 = 20
\(x\) - 25 = 20 x 15
\(x\) - 25 = 300
\(x\) = 300 + 25
\(x\) = 325
c, (3\(x\) - 15).7 = 42
3\(x\) - 15 = 42:7
3\(x\) - 15 = 6
3\(x\) = 6 + 15
3\(x\) = 21
\(x\) = 21: 3
\(x\) = 7
Ta có : \(\frac{x-99}{5}+\frac{x-99}{15}+\frac{x-99}{25}+\frac{x-99}{35}=0\)
\(\Rightarrow\left(x-99\right)\left(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+\frac{1}{35}\right)=0\)
Vì \(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+\frac{1}{35}\ne0\)
Nên : x - 99 = 0
Suy ra : x = 99
Vậy x = 99