Câu hỏi của Nguyễn Tú Anh

Question

$\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2$ phân tích đa thức thành nhân tử

Bùi Mạnh Khôi · Accepted Answer

$\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2$

$=\left(x^2-x+1\right)^2-2\left(x^2-x+1\right)\dfrac{5}{2}x+\left(\dfrac{5}{2}x\right)^2-\left(\dfrac{3}{2}x\right)^2$

$=\left(x^2-x+1-\dfrac{5}{2}x\right)^2-\left(\dfrac{3}{2}x\right)^2$

$=\left(x^2-\dfrac{7}{2}x+1\right)^2-\left(\dfrac{3}{2}x\right)^2$

$=\left(x^2-\dfrac{7}{2}x+1-\dfrac{3}{2}x\right)\left(x^2-\dfrac{7}{2}x+1+\dfrac{3}{2}x\right)$

$=\left(x^2-5x+1\right)\left(x^2-2x+1\right)$Câu trả lời: $\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2$

$=\left(x^2-x+1\right)^2-2\left(x^2-x+1\right)\dfrac{5}{2}x+\left(\dfrac{5}{2}x\right)^2-\left(\dfrac{3}{2}x\right)^2$

$=\left(x^2-x+1-\dfrac{5}{2}x\right)^2-\left(\dfrac{3}{2}x\right)^2$

$=\left(x^2-\dfrac{7}{2}x+1\right)^2-\left(\dfrac{3}{2}x\right)^2$

$=\left(x^2-\dfrac{7}{2}x+1-\dfrac{3}{2}x\right)\left(x^2-\dfrac{7}{2}x+1+\dfrac{3}{2}x\right)$

$=\left(x^2-5x+1\right)\left(x^2-2x+1\right)$

Violympic toán 8