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20 tháng 10 2021

\(\left(4x-3\right)\left(3x+2\right)-\left(6x+1\right)\left(2x+5\right)+1\)

\(=\left(12x^2-9x+8x-6\right)-\left(12x^2+2x+30x+5\right)+1\)

\(=\left(-x-32x\right)+\left(-6-5+1\right)=-33x-10\)

10 tháng 10 2018

\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^{128}-1\right)< B\)

10 tháng 10 2018

\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)

\(\Rightarrow A< B\)

29 tháng 7 2021

ý bạn là nhân đa thức với đa thức hay sao ạ?

29 tháng 6 2023

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

9 tháng 8 2016

a) \(A=1999\cdot2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)

=> \(A< B\)

b) \(A=12^6\)

    \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

       \(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

      \(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

      \(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

      \(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)

=> \(A>B\)

c) \(A=2011\cdot2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)

   \(B=2012^2\)

=> \(A< B\)

d) \(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

        \(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)

          \(=\frac{\left(3^4-1\right)\left(3^4+1\right)..\left(3^{64}+1\right)}{2}\)

          \(=\frac{\left(3^8-1\right).....\left(3^{64}+1\right)}{2}\)

           \(=\frac{3^{128}-1}{2}\)

 \(B=3^{128}-1\)

=> \(A< B\)

Cảm ơn bạn 

16 tháng 9 2021

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3