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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)
\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)
\(A=\left(-2x-4\right)^2\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)