Question

Phân tích A=(x-2)(x-4)(x-6)(x-8)+4 thành nhân tử

Answer 1

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+4\\ =\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+4\\ =\left(x^2-10x+16\right)\left(x^2-10x+24\right)+4\\ t=x^2-10x+20\\ \Rightarrow\left(t+4\right)\left(t-4\right)+4\\ =t^2-16+4\\ =t^2-12\\ =\left(t-\sqrt{12}\right)\left(t+\sqrt{12}\right)\\ =\left(x^2+10x+20-\sqrt{12}\right)\left(x^2+10x+20+\sqrt{12}\right)\)

( phân tích trong ngoặc ra)