Bài này đơn giản như đang dỡn
\(x^4+4=x^4+4x^2+4-4x^2\)![]()
\(=\left(x^2+2\right)^2-\left(2x\right)^2\) ![]()
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\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)![]()
phân tích thành đa nhân tử?
\(x^4+4=\left(x^2\right)^2+2^2\)
\(=\left(x^2\right)^2+4x^2+2^2-4x^2\)
\(=\left[\left(x^2\right)^2+4x^2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
\(x^4+4=\left(x^2\right)^2-4x^2+4-4x^2=\left[\left(x^2\right)^2-4x^2+4\right]-\left(2x\right)^2=\left(x^2-2\right)^2-\left(2x\right)^2=\left(x^2+2x-2\right)\left(x^2-2x-2\right)\)
