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Bài 4 : Tính nhanh :
a, 15. 64 + 25. 100 + 36. 15 + 60. 100
= (15 . 64 + 36. 15) + (25. 100 + 60. 100)
= 15.(64 + 36) + 100.(25 + 60)
= 15. 100 + 100. 85
= 100.(15 + 85)
= 100. 100
= 10000
b, 472 + 482 - 25 + 94. 48
= 472 + 2.47. 48 + 482 - 25
= (47 + 48)2 - 52
= (47 + 48 - 5)(47 + 48 + 5)
= (48 + 22)(48 + 52)
= 90. 100
= 9000
c, 93 - 92. ( -1) - 9. 11 + ( -1). 11
= 93 + 92 + 11(- 9 - 1)
= 92.(9 + 1) + 11. (-10)
= 81. 10 - 110
= 810 - 110
= 700
d,2016. 2018 - 20172
= (2017 - 1)(2017 + 1) - 20172
= 20172 - 1 - 20172
= -1
#Học tốt!
tách ít ít ra thôi. để cả cộp thế này k ai làm cho đâu. mệt quá
a) \(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)
b) \(\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{64}-1\right)\)
\(=\dfrac{9^{64}-1}{10}\)
Ta có: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)
Mà \(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)
\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)
c) Ta có:
\(\dfrac{x^2-y^2}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-xy}\left(1\right)\)
Vì x>y>0, ta có:
\(\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)
Vì x>y>0 nên \(\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)\)
Từ (1), (2) và (3) suy ra:
\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)
a) Ta có:
\(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
Vậy 2011.2013+2012.2014 = 20122 + 20132 - 2
Ta có \(\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\frac{1}{8}\left(9-1\right)\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\frac{1}{8}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
cứ như thế
\(=\frac{1}{8}\left(9^{64}-1\right)< 9^{64}-1\)=>đpcm
Ta có A = 2018.2020 + 2019.2021
= (2020 - 2).2020 + 2019.(2019 + 2)
= 20202 - 2.2020 + 20192 + 2.2019
= 20202 + 20192 - 2(2020 - 2019) = 20202 + 20192 - 2 = B
=> A = B
b) Ta có B = 964 - 1= (932)2 - 12
= (932 + 1)(932 - 1) = (932 + 1)(916 + 1)(916 - 1) = (932 + 1)(916 + 1)(98 + 1)(98 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(94 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(92 - 1)
(932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).80
mà A = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).10
=> A < B
c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)
=> A < B
d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)
=> A < B
ta có 2015 x 2017 >2017^2 -2
2016 x 2018 > 2016^2
=> A> B