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Nhân hết ra rồi rút gọn thôi bạn:
\(A=\left(2^9+2^7+1\right)\left(2^{23}-2^{21}+2^{19}-2^{17}+2^{14}-2^{10}+2^9-2^7+1\right).\)
\(=2^{32}-2^{30}+2^{28}-2^{26}+2^{23}-2^{19}+2^{18}-2^{16}+2^9\)\(+2^{30}-2^{28}+2^{26}-2^{24}+2^{21}-2^{17}+2^{16}-2^{14}+2^7+2^{23}-2^{21}+2^{19}-2^{17}+2^{14}-2^{10}\)\(+2^9-2^7+1\)
\(=2^{32}+\left(2^{23}+2^{23}-2^{24}\right)+\left(2^{18}-2^{17}-2^{17}\right)+\left(2^9+2^9-2^{10}\right)+1=2^{32}+1\)
A = 232 + ( 223 + 223-224) + (218 - 217 - 217) + ( 29 + 29 - 210) + 1
= 223 + 1
a) \(A=85^2-45^2+75^2-35^2+65^2-25^2+55^2-15^2\)
\(A=\left(85-45\right)\left(85+45\right)+....+\left(55-15\right)\left(55+15\right)\)
\(A=40.130+40.110+40.90+40.70\)
\(A=40.\left(130+110+90+70\right)=40.400=16000\)
b) \(B=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2011-2012\right)\left(2011+2012\right)\)
\(B=-3-7-11-...-4023\)
\(B=-\left(3+7+11+...+4023\right)\)
\(B=-\dfrac{\left(3+4023\right)\left[\dfrac{\left(4023-3\right)}{4}+1\right]}{2}=2025078\)
\(A=2^{32}-2^{30}+2^{28}-2^{26}+2^{23}-2^{19}+2^{18}-2^{16}+2^9\)
\(+2^{30}-2^{28}+2^{26}-2^{24}+2^{21}-2^{17}+2^{16}-2^{14}+2^7\)
\(+2^{23}-2^{21}+2^{19}-2^{17}+2^{14}-2^{10}+2^9-2^7+1\)
\(=2^{32}+1\)
Bài này khi nhận thông thường thì ta rút gọn đc hết. :)
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