12.102 - ( 15.102.2+18.102.2.3 )+2.102 = ?
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a) ( 100 – 1 ) 3 = 970299. b) ( 91 + 9 ) 3 = 100 3 .
c) ( 1000 + 1 ) 3 = 1003003001. d) ( 102 – 2 ) 3 = 100 3 .
\(\left(7x-11\right)^3=2^5.5^2+2.10^2\)
\(\Rightarrow\left(7x-11\right)^3=32.25+2.100\)
\(\Rightarrow\left(7x-11\right)^3=800+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
(7\(x\) - 11)3 = 25.52 + 2.102
(7\(x\) - 11)3 = 1000
(7\(x\) - 11)3 = 103
7\(x\) - 11 = 10
7\(x\) = 10 + 11
7\(x\) = 21
\(x\) = 21:7
\(x\) = 3
\(\dfrac{101}{2}.\dfrac{102}{2}.\dfrac{103}{2}.\dfrac{104}{2}.....\dfrac{200}{2}\\ =\dfrac{101.102.103.104.....200}{2^{100}}\\ =\dfrac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}.\left(1.2.3.....100\right)}\\ =\dfrac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right).....\left(2.100\right)}\\ =\dfrac{\left(1.3.5.....199\right)\left(2.4.6.....200\right)}{4.6.8.....200}\\ =1.3.5.7.....197.199\)
=> Điều phải chứng minh
\(\frac{101}{2}\times\frac{102}{2}\times\frac{103}{2}\times...\times\frac{200}{2}\)
\(=\frac{1.2.3.....100.101.102.103.....200}{1.2.3.....100.2^{100}}\)
\(=\frac{\left(1.3.5.....199\right).\left(2.4.6.....200\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}\)
\(=1.3.5.....199\)
Ta cói: 12.10^2 - (15.10^2.2+18.10^2.2.3) + 2.10^2
= 12.10^2 - [10^2.(15.2-18.2.3)] + 2.10^2
= 12.10^2 - 10^2.(-78)+2.10^2
=10^2.(12-(-78)+2)
=10^2.(12+78-2)
=10^2.88
=100.88
=8800
\(12.10^2-\left(15.10^2.2+18.10^2.2.3\right)+2.10^2\)
\(=1200-\left(3000+10800\right)+2.10^2\)
\(=1200-13800+200\)
\(=-12600+200=-12400\)