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\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) 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: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

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

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

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

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

26 tháng 7 2021

a) A=(4-5x)2-(3+5x)2=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

26 tháng 7 2021

B=(3x-1)(1+3x)-(3x+1)2=9x2-1-(3x+1)2=9x2-1-(9x2+6x+1)=9x2-1-9x2-6x-1=-6x-2=-2(3x+1)

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

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

\(A=\left(-2x-4\right)^2\)

30 tháng 9 2017

A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2

= [(3x + 1)-(5x + 5)]2

= (3x + 1 - 5x - 5)2

= [(-2x) - 4]2

B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)

= (38 - 1)(38 + 1)(316 +1)(332 + 1)

= (316 - 1)316 +1)(332 + 1)

= (332 - 1)(332 + 1)

= 364 - 1

vì 2B = 364 - 1

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

C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)

= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2

= 2a2