Câu hỏi của thùy linh

Question

bài 5.phân tích đa thức sau thành nhân tử17,$x^3-2x^2+x$18,$3\left(x+4\right)-x^2-4x$19,$x^2+5x-6$20,$x^2+x-20$

Ngô Hải Nam · Accepted Answer

17)$x^3-2x^2+x\
=x\left(x^2-2x+1\right)\
=x\left(x-1\right)^2$18)$3\left(x+4\right)-x^2-4x\
=3\left(x+4\right)-x\left(x+4\right)\
=\left(x+4\right)\left(3-x\right)$19)$x^2+5x-6\
=x^2+6x-x-6\
=x\left(x+6\right)-\left(x+6\right)\
=\left(x+6\right)\left(x-1\right)$20)$x^2+x-20\
=x^2+5x-4x-20\
=x\left(x+5\right)-4\left(x+5\right)\
=\left(x+5\right)\left(x-4\right)$Câu trả lời: 17)$x^3-2x^2+x\
=x\left(x^2-2x+1\right)\
=x\left(x-1\right)^2$18)$3\left(x+4\right)-x^2-4x\
=3\left(x+4\right)-x\left(x+4\right)\
=\left(x+4\right)\left(3-x\right)$19)$x^2+5x-6\
=x^2+6x-x-6\
=x\left(x+6\right)-\left(x+6\right)\
=\left(x+6\right)\left(x-1\right)$20)$x^2+x-20\
=x^2+5x-4x-20\
=x\left(x+5\right)-4\left(x+5\right)\
=\left(x+5\right)\left(x-4\right)$

#Blue Sky · Answer

$17,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2$$18,3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)$$19,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)$$20,x^2+x-20=x^2-4x+5x-20=x\left(x-4\right)+5\left(x-4\right)=\left(x-4\right)\left(x+5\right)$Câu trả lời: $17,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2$$18,3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)$$19,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)$$20,x^2+x-20=x^2-4x+5x-20=x\left(x-4\right)+5\left(x-4\right)=\left(x-4\right)\left(x+5\right)$

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)