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\(\left(a^2+b^2-c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2\right)^2-\left(2ab\right)^2\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[\left(a-b\right)^2+c^2\right]\)
=(a+b+c)(a+b-c)(a-b+c)(a-b-c)
a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)
b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)
c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)
d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)
a) \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=\left(x^2+x\right)+\left(3x+3\right)\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
c) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
\(=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-4c^2\)
\(=2\left(a+b\right)^2-2c^2=2\left[\left(a+b\right)^2-c^2\right]=2\left(a+b+c\right)\left(a+b-c\right)\)
Mình đã làm bài này rồi.
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\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2\)
\(=\left[\left(a+b+c\right)^2-\left(2c\right)^2\right]+\left(a+b-c\right)^2\)
\(=\left(a+b+3c\right)\left(a+b-c\right)+\left(a+b-c\right)^2\)
\(=\left(a+b-c\right)\left(a+b+3c+a+b-c\right)\)
\(=\left(a+b-c\right)\left(2a+2b+2c\right)\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)