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Question

Phân tích thành đa nhân tử

$\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1$

$\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16$

Linh Nguyễn · Accepted Answer

$A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1$

$A=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1$

$A=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]+1$

$A=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)+1$

$A=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1$

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

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

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

$B=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16$

$B=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16$

5Câu trả lời: $A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1$