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\(53.\left(51+4\right)+53.\left(49+91\right)+53\\ =53.55+53.140+53.1\\ =53.\left(55+140+1\right)\\ =53.196\\ =10388.\)
\(53\cdot\left(51+4\right)+\left(49+91\right)+53\)
\(=53\cdot55+53\cdot140+53\cdot1\)
\(=53\cdot\left(55+140+1\right)\)
\(=53\cdot196\)
\(=53\cdot\left(200-4\right)\)
\(=53\cdot200+53\cdot4\)
\(=10600-212\)
\(=10388\)
\(53.51+4+53.49+91+53\)
\(=53.\left(51+49+1\right)+4+91\)
\(=53.101+4+91\)
\(=5353+4+91\)
\(=5448\)
53.51+51.4+ 53.49+ 53.96+53
= 53.(51+49)+ 53.( 96+4)+ 53
= 53.100+ 53. 100 + 53
= 5300+5300+53
=10600+53
=10653
****
\(53\cdot\left(51+4\right)+53\cdot\left(49+96\right)+53\)
=\(53\cdot55+53\cdot145+53\cdot1\)
=\(53\cdot\left(55+145+1\right)\)
=\(53\cdot201\)
=10653
nếu Đ thì k cho mk nhá
`#3107.101107`
\(\text{53 x 51 + 4 + 53 x 49 + 91 + 53}\)
\(=53\times\left(51+49\right)+\left(4+91+53\right)\\ =53\times100+148\\ =5300+148\\ =5448\)
\(53\times51+4+53\times49+91+53\)
\(=53\times\left(51+49\right)+\left(4+53+91\right)\)
\(=53\times100+148\)
\(=5300+148\)
\(=5448\)
`47/53*(17/3-53/47)+17/3*(6/17-47/53)`
`=47/53*17/3-1+6/3-17/3*47/53`
`=2-1+17/3*47/53-17/3*47/53`
`=1`
\(\dfrac{47}{53}.\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}.\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\)
\(=\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}-\dfrac{17}{3}.\dfrac{47}{53}\)
\(=\left(\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{17}{3}.\dfrac{47}{53}\right)+\left(\dfrac{-47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}\right)\)
\(=\left(1-1\right)+\left(-1+2\right)\)
\(=0+1\)
\(=1\)
Ta có: \(\dfrac{11}{53}+\left(\dfrac{32}{47}+\dfrac{-10}{53}\right)+\dfrac{-64}{94}+\dfrac{1}{-53}+\dfrac{1}{3}\)
\(=\dfrac{11}{53}+\dfrac{32}{47}-\dfrac{10}{53}-\dfrac{32}{47}-\dfrac{1}{53}+\dfrac{1}{3}\)
\(=\dfrac{1}{3}\)
n!=1x 2x3 x4 x5 x..........xn
=>53!= 1 x 2 x 3 x 4 x 5 x....................x 53
4.274883284x1069