Ta có:
\(x^4+4\)
\(=\left(x^4+4x^2+4\right)-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
Áp dụng hằng đẳng thức:\(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
có:\(\left(x^2+2\right)^2-\left(2x\right)^2\)
=\(\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
x4 +4
= x4 +4x2 -4x2 +4
= (x4+4x2+4)-4x2
= (x2+2)2-4x2
= \((x^2+2-2x)(x^2+2+2x)\)
Ta có: x4+4= (x4+4x2+4)-4x2
= (x2+2)2 - 4x2
= (x2+2+4x).(x2+2-4x)
\(x^4+4\\ \\=x^4+4+4x^2-4x^2\\ \\=\left(x^4+4x^2+4\right)-4x^2\\ \\=\left(x^2+2\right)^2-\left(2x\right)^2\\ \\=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
