Bài 2 phân tích đa thức thành nhân tử
a,x^2(x-1)+16(1-x)
b,5x^2-5y^2-10x+10y
c,x^2+4x-y^2+4
K
Khách
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13 tháng 8 2018
1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)
3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)
5) \(x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)=\left(x-1\right)^2\left(x+1\right)\)
6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
BM
1
TD
1
KN
29 tháng 10 2019
a) \(x^2-5xy+6y^2\)
\(=x^2-3xy-2xy+6y^2\)
\(=x\left(x-3y\right)-2y\left(x-3y\right)\)
\(=\left(x-2y\right)\left(x-3y\right)\)
b) \(16\left(x-1\right)^2-36y^2\)
\(=\left(4x-4\right)^2-\left(6y\right)^2\)
\(=\left(4x+6y-4\right)\left(4x-6y-4\right)\)
c) \(4\left(x+y\right)-12\left(x+y\right)^2\)
\(=\left(x+y\right)\left[4-12\left(x+y\right)\right]\)
\(=4\left(x+y\right)\left[1-3x-3y\right]\)
LN
1
LH
1
30 tháng 10 2018
\(5x^2-x+y-5y^2\)
\(=\left(5x^2-5y^2\right)-\left(x-y\right)\)
\(=5\left(x^2-y^2\right)-\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left[5\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(5x+5y-1\right)\)
TD
1
4 tháng 7 2017
\(=\left(x^2+4x-3\right)^2-5\left(x^2+4x-3\right)+6x^2\)
\(=x^4+16x^2+9+8x^3-24x-6x^2-5x^2-20x+15+6x^2\)
\(=x^4+8x^3+11x^2-44x+24\)
\(=\left(x^4-x^3\right)+\left(9x^3-9x^2\right)+\left(20x^2-20x\right)-\left(24x-24\right)\)
\(=x^3\left(x-1\right)+9x^2\left(x-1\right)+20x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+9x^2+20x-24\right)\)
12 tháng 12 2021
\(=x^2-2xy-5xy+10y^2=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-2y\right)\left(x-5y\right)\)
DN
3
2 tháng 9 2021
\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
a) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)
\(=x^2\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-16\right)\)
\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
b) Ta có: \(5x^2-5y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\cdot2\)
\(=5\left(x-y\right)\left(x+y-2\right)\)
c) Ta có: \(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)