Câu hỏi của Đào Nguyễn Hồng Ngọc

Question

Phân tích đa thức thành nhân tử:a) (x2 +x)2 - 14(x2 +x) + 24b) (x2 +x)2 + 4x2 4x - 12c) (x+1) (x+2) (x+3) (x+4) + 1d) (x+1) (x+3) (x+5) (x+7) + 15e) x4 + 2x3 +5x2 +4x - 12f) (x2 + x + 1)(x2 +x +2) - 1...

Phương An · Accepted Answer

$A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24$Đặt $x^2+x=t$, ta có:$A=t^2-14t+24$$=t^2-2t-12t+24$$=t\left(t-2\right)-12\left(t-2\right)$$=\left(t-2\right)\left(t-12\right)$$=\left(x^2+x-2\right)\left(x^2+x-12\right)$$B=\left(x^2+x\right)^2+4x^2+4x-12$$=\left(x^2+x\right)^2+4\left(x^2+x\right)-12$Đặt $x^2+x=t$, ta có:$B=t^2+4t-12$$=t^2+6t-2t-12$$=t\left(t+6\right)-2\left(t+6\right)$$=\left(t+6\right)\left(t-2\right)$$=\left(x^2+x+6\right)\left(x^2+x-2\right)$$C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1$$=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1$Đặt $x^2+5x+4=t$, ta có:$C=t\left(t+2\right)+1$$=t^2+2t+1$$=\left(t+1\right)^2$$=\left(x^2+5x+4+1\right)^2$$=\left(x^2+5x+5\right)^2$$D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15$$=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15$Đặt $x^2+8x+7=t$, ta có:$D=t\left(t+8\right)+15$$=t^2+8t+15$$=t^2+3t+5t+15$$=t\left(t+3\right)+5\left(t+3\right)$$=\left(t+3\right)\left(t+5\right)$$=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)$$=\left(x^2+8x+10\right)\left(x^2+8x+12\right)$$F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12$Đặt $x^2+x+1=t$, ta có:$F=t\left(t+1\right)-12$$=t^2+t-12$$=t^2+4t-3t-12$$=t\left(t+4\right)-3\left(t+4\right)$$=\left(t+4\right)\left(t-3\right)$$=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)$$=\left(x^2+x+5\right)\left(x^2+x-2\right)$$E=x^4+2x^3+5x^2+4x-12$$=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12$$=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)$$=\left(x-1\right)\left(x^3+3x^2+8x+12\right)$$=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)$$=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]$$=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)$ Câu trả lời: $A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24$Đặt $x^2+x=t$, ta có:$A=t^2-14t+24$$=t^2-2t-12t+24$$=t\left(t-2\right)-12\left(t-2\right)$$=\left(t-2\right)\left(t-12\right)$$=\left(x^2+x-2\right)\left(x^2+x-12\right)$$B=\left(x^2+x\right)^2+4x^2+4x-12$$=\left(x^2+x\right)^2+4\left(x^2+x\right)-12$Đặt $x^2+x=t$, ta có:$B=t^2+4t-12$$=t^2+6t-2t-12$$=t\left(t+6\right)-2\left(t+6\right)$$=\left(t+6\right)\left(t-2\right)$$=\left(x^2+x+6\right)\left(x^2+x-2\right)$$C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1$$=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1$Đặt $x^2+5x+4=t$, ta có:$C=t\left(t+2\right)+1$$=t^2+2t+1$$=\left(t+1\right)^2$$=\left(x^2+5x+4+1\right)^2$$=\left(x^2+5x+5\right)^2$$D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15$$=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15$Đặt $x^2+8x+7=t$, ta có:$D=t\left(t+8\right)+15$$=t^2+8t+15$$=t^2+3t+5t+15$$=t\left(t+3\right)+5\left(t+3\right)$$=\left(t+3\right)\left(t+5\right)$$=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)$$=\left(x^2+8x+10\right)\left(x^2+8x+12\right)$$F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12$Đặt $x^2+x+1=t$, ta có:$F=t\left(t+1\right)-12$$=t^2+t-12$$=t^2+4t-3t-12$$=t\left(t+4\right)-3\left(t+4\right)$$=\left(t+4\right)\left(t-3\right)$$=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)$$=\left(x^2+x+5\right)\left(x^2+x-2\right)$$E=x^4+2x^3+5x^2+4x-12$$=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12$$=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)$$=\left(x-1\right)\left(x^3+3x^2+8x+12\right)$$=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)$$=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]$$=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)$

Đặng Quỳnh Ngân · Answer

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