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Bài 1:
a) \(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
b) Sửa đề \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)-3^{64}\)
\(=3^{64}-1-3^{64}\)
\(=-1\)
Bài 2:
Ta có:
\(A=2009.2009\)
\(A=2009\left(2008+1\right)\)
\(A=2009.2008+2009\)
Ta lại có:
\(B=2008.2010\)
\(B=2008\left(2009+1\right)\)
\(B=2008.2009+2008\)
Vì 2008.2009 = 2009.2008
2009 > 2008
=> 2008.2009 + 2009 > 2009.2008 + 2008
=> A > B
1,a,(2-1)(2+1)(22+1)(24+1)(28+1)
=(22-1)(22+1)(24+1)(28+1)
=(24-1) (24+1)(28+1)
=(28 -1)(28+1)=216-1
2,
A=2009.2009=20092
B=2008.2010=(2009-1)(2009+1)=20092-1
Do20092>20092-1\(\Rightarrow A>B\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}=B\)
b) A = 2010 . 2012
= ( 2011 - 1 )( 2011 + 1 )
= 20112 - 12 = 20112 - 1
20112 - 1 < 20112 => A < B
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
a) Đặt \(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=2.\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^4-1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(...\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2A=3^{64}-1\)
\(A=\frac{3^{64}-1}{2}\)
a) \(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=.............................................................\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1=B-1\)
Suy ra A < B
b) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1=B-1\)
Suy ra A < B
Phần a bạn nhân thêm ở A là (2-1) là ra hằng đẳng thức, cứ thế mà triển. (Kết quả: A<B)
Phần b: phân tích A, ta có:
2015.2017= (2016-1).(2016+1)= 2016^2 -1 <2016^2
Suy ra: A<B
Bài 3:
a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)
\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)
\(\Leftrightarrow x^3-27-x^3+16x=21\)
=>16x=48
hay x=3
b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3+8-x^3-2x=4\)
=>-2x=4-8=-4
hay x=2
\(a)\) Ta có :
\(A=2008.2010=\left(2009+1\right)\left(2009-1\right)=2009^2-1< 2009^2=B\)
Vậy \(A< B\)
\(b)\)\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\) (bn xem lại đề xem có nhầm j ko, nếu đề đúng thì mk sr)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(A=2^{32}-1< 2^{32}=B\)
Vậy \(A< B\)
Chúc bạn học tốt ~