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5:

a: (2x-5)(2x+5)=4x^2-25

b: (3x-5y)(3x+5y)=9x^2-25y^2

c: (3x+7y)(3x-7y)=9x^2-49y^2

d: (2x-1)(2x+1)=4x^2-1

4:

a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2

b: 8(7^2+1)(7^4+1)(7^8+1)

=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)

=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)

=1/6(7^16-1)<7^16-1

28 tháng 7 2023

5:

a: (2x-5)(2x+5)=4x^2-25

b: (3x-5y)(3x+5y)=9x^2-25y^2

c: (3x+7y)(3x-7y)=9x^2-49y^2

d: (2x-1)(2x+1)=4x^2-1

mik chỉ biết bài 5 thôi !

1 tháng 10 2016

Bài 1 :

(a^2+b^2)(x^2+y^2)=(ax+by)^2 
<=> a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2 
<=> a^2y^2 + b^2x^2 = 2abxy 
<=> a^2y^2 + b^2x^2 - 2abxy = 0 
<=> (ay - bx)^2 = 0 
=> ay - bx = 0 
=> ay = bx 
=> a/x = b/y ( x,y khác 0)

 

 

\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\) =\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\) =25 c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\) =\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\) =\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\) =... =\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\) \)

d)Tương tự

\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

=\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\)

=\(a^2\)

b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

=\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\)

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

=\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\)

=25

c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\)

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

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

=...

=\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)

d)Tương tự

5 tháng 9 2017

cảm ơn

a: \(=\left[a-\left(b-c\right)\right]^2-\left(b-c\right)^2+2ab-2ac\)

\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2ab-2ac\)

\(=a^2-2ab+2ac+2ab-2ac=a^2\)

b: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left(3x+1-3x-5\right)^2\)

\(=\left(-4\right)^2=16\)

c: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)

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

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

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

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(=2^{128}-1\)

d: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)

\(=\dfrac{3^{64}-1}{2}\)

2 tháng 10 2018

1)a)=>x2+y2+2xy-4(x2-y2-2xy)

=>x2+y2+2xy-4.x2+4y2+8xy

=>-3.x2+5y2+10xy