Tính nhanh :
2^3 .3^2.28+2^3.3^5.8
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c) Ta có: x+11 chia hết cho x+1
x+1+10 chia hết cho x+1
Vì x+1 chia hết cho x+1 nên 10 chia hết cho x+1.
Vậy x+1\(\inƯ\left(10\right)=\left\{1;2;5;10\right\}\)
\(\Rightarrow x\in\left\{0;1;4;9\right\}\)
\(25.2^3+75.2^3=2^3.\left(25+75\right)=8.100=800\)
\(3^2.187-87.3^2=3^2.\left(187-87\right)=9.100=900\)
\(4^2.19+80.4^2+4^2=4^2\left(19+80+1\right)=16.100=1600\)
\(73.5^2+5^2.28-5^2=5^2\left(73+28-1\right)=25.100=2500\)
\(6^2.48+51.6^2+36=6^2\left(48+51+1\right)=36.100=3600\)
\(113.7^2-7^2.12-49=49.\left(113-12-1\right)=4900\)
a. 25.23 + 75.23
= 23(25 + 75)
= 8 . 100
= 800
b. 32 . 187 - 87.32
= 32 . (187 - 87)
= 9 . 100
= 900
c. 42 . 19 + 80 . 42 + 42
= 42 . (19 + 80 + 1)
= 16 . 100
= 1600
d. 73 . 52 + 52. 28 - 52
= 52 . (73 + 28 - 1)
= 25 . 100
= 2500
e. 62 . 48 + 51 . 62 + 36
= 36 . 48 + 51 . 36 + 36
= 36 . (48 + 51 + 1)
= 36 . 100
= 3600
g. 113.72 - 72.12 - 49
= 113 . 49 - 49 . 12 - 49
= 49. (113 - 12 - 1)
= 49 . 100
= 4900
j. (23 . 94 + 93 . 45) : (92 . 10 - 92)
= (23 . 94 + 93 . 9.5) : [92(10 - 1)]
= [9. (23 . 93 + 92 . 9.5)] : (92 . 9)
= {9. [92 (23 + 9.5)]} : (92 . 9)
= (9. 92 . 53): 93
= 93 . 53 : 93
= (93:93) . 53
= 53
\(A=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}\right)=3.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=3.\left(\frac{11}{22}-\frac{2}{22}\right)=3.\frac{9}{22}=\frac{27}{22}\)
3/2.5 + ...+ 3/17 .20
= 3/2 .(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 120)
= 3/2 . (1/2 - 1/20)
= \(\frac{3}{2}\) . \(\frac{9}{20}\) = \(\frac{27}{40}\)
\(2^3.3^2.28+2^3.3^5.8\\ =2^3.3^2.2^2.7+2^3.3^5.2^3\\ =2^5.3^2.7+2^6.3^5\\ =2^5.3^2\left(7+2.3^3\right)\\ =32.9\left(7+54\right)\\ =32.9.61=17568\)
\(2^3.3^2.28+2^3.3^5.28\)
=> \(28.(2^3.3^2+2^3.3^5)\)
=> \(28.[3^3(3^2+3^5)]\)
=> \(28.[3^3.252]\)
=> ............( tự tính)