Câu hỏi của Thu Theo

Question

1/ x2 +x-6

2/x2 -3x +2

3/x2 -4x+3

Nguyễn Huy Tú · Accepted Answer

1, $x^2+x-6=x^2+3x-2x-6$

$=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)$

2, $x^2-3x+2=x^2-2x-x+2$

$=x\left(x-2\right)-\left(x-2\right)=\left(x-1\right)\left(x-2\right)$

3, $x^2-4x+3=x^2-3x-x+3$

$=x\left(x-3\right)-\left(x-3\right)=\left(x-1\right)\left(x-3\right)$Câu trả lời: 1, $x^2+x-6=x^2+3x-2x-6$

$=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)$

2, $x^2-3x+2=x^2-2x-x+2$

$=x\left(x-2\right)-\left(x-2\right)=\left(x-1\right)\left(x-2\right)$

3, $x^2-4x+3=x^2-3x-x+3$

$=x\left(x-3\right)-\left(x-3\right)=\left(x-1\right)\left(x-3\right)$

Trần Đăng Nhất · Answer

1) $x^2+x-6$

$=x^2+3x-2x-6$

$=x\left(x-3\right)-2\left(x-3\right)$

$=\left(x-3\right)\left(x-2\right)$

2) $x^2-3x+2$

$=x^2+4x-x+2$

$=x\left(x+2\right)-\left(x+2\right)$

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

3) $x^2-4x+3$

$=x^2-3x-x+3$

$=x\left(x-3\right)-\left(x-3\right)$

$=\left(x-3\right)\left(x-1\right)$Câu trả lời: 1) $x^2+x-6$