phân tích thành nhân tử 1) x^2 – 7xy + 10y^2 2) x^3 – 5x^2 – 14x 3) x^2.yz + 5xyz – 14yz 4) a^4 + 4a^2 – 5 5) x^3 + 4x^2 – 29x +24

0 bình luận về “phân tích thành nhân tử 1) x^2 – 7xy + 10y^2 2) x^3 – 5x^2 – 14x 3) x^2.yz + 5xyz – 14yz 4) a^4 + 4a^2 – 5 5) x^3 + 4x^2 – 29x +24”

1. $1)_{}$ $x^2-7xy+10y^2_{}$

   $⇔x^2-5xy-2xy+10y^2_{}$

   $⇔x.(x-5y)-2y.(x-5y)_{}$

   $⇔(x-2y)(x-5y)_{}$

   $2)_{}$ $x^3-5x^2-14x_{}$

   $⇔x^3+2x^2-7x^2-14x_{}$

   $⇔x^2.(x+2)-7x.(x+2)_{}$

   $⇔(x^2-7x).(x+2)_{}$

   $⇔x.(x-7)(x+2)_{}$

   $3)_{}$ $x^2yz+5xyz-14yz_{}$

   $⇔yz.(x^2+5x-14)_{}$

   $⇔yz.(x^2+7x-2x-14)_{}$

   $⇔yz.[ x.(x+7)-2.(x+7)]_{}$

   $⇔yz.(x+7).(x-2)_{}$

   $5)_{}$ $x^3+4x^2-29x+24_{}$

   $⇔x^3-x^2+5x^2-5x-24x+24_{}$

   $⇔x^2.(x-1)+5x.(x-1)-24.(x-1)_{}$

   $⇔(x-1).(x^2+5x-24)_{}$

   $⇔(x-1)(x^2+8x-3x-24)_{}$

   $⇔(x-1). [ x.(x+8)-3.(x+8)]_{}$

   $⇔(x-1)(x+8)(x-3)_{}$

   $4)_{}$ $a^4+4a^2-5_{}$

   $⇔a^4+5a^2-a^2-5_{}$

   $⇔a^2.(a^2+5)-(a^2+5)_{}$

   $⇔(a^2+5)(a^2-1)_{}$

   $⇔(a^2+5)(a-1)(a+1)_{}$

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2. $1)x²-7xy+10y²$

   $=x²-2xy-5xy+10y²$

   $=x(x-2y)-5y(x-2y)$

   $=(x-5y)(x-2y)$

   $2)x³-5x²-14x$

   $=x³+2x²-7x²-14x$

   $=x²(x+2)-7x(x+2)$

   $=(x²-7x)(x+2)$

   $=x(x-7)(x+2)$

   $3)x²yz+5xyz-14yz$

   $=x²yz-2xyz+7xyz-14yz$

   $=xyz(x-2)+7yz(x-2)$

   $=(xyz+7yz)(x-2)$

   $=yz(x+7)(x-2)$

   $4)a^4+4a²-5$

   $=a^4-a²+5a²-5$

   $=a²(a²-1)+5(a²-1)$

   $=(a²+5)(a²-1)$

   $=(a²+5)(a-1)(a+1)$

   $5)x³+4x²-29x+24$

   $=x³-x²+5x²-5x-24x+24$

   $=x²(x-1)+5x(x-1)-24(x-1)$

   $=(x²+5x-24)(x-1)$

   $=(x²+8x-3x-24)(x-1)$

   $=[x(x+8)-3(x+8)](x-1)$

   $=(x-3)(x+8)(x-1)$

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