phan tich da thuc thanh nhan tu
a: 4x^2-20xy +25-[3x-2]^2
b: x^2-6x+5+[x-5]^2
phan tich da thuc thanh nhan tu
A=x^6-2x^5-4x^4+6x^3+4x^2-2x-1
Phan tich da thuc thanh nhan tu
3x^2-11x+6
x^2-6x+5
x^4+x^2+1
x^4-4x^2+3
6x^2+7xy+2y^2
(*)\(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(x-3\right)\left(3x-2\right)\)
(*)\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-5\right)\left(x-1\right)\)
(*)\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+1+x\right)\left(x^2+1-x\right)\)
(*)\(x^4-4x^2+3=x^4-x^2-3x^2+3=x^2\left(x^2-1\right)-3\left(x^2-1\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
(*)\(6x^2+7xy+2y^2=6x^2+4xy+3xy+2y^2=2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(2x+y\right)\left(3x+2y\right)\)
a, \(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(x-3\right)\)
b, \(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
c, \(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
d, \(x^4-4x^2+3=x^4-4x^2+4-1=\left(x^2-2\right)^2-1=\left(x^2-1\right)\left(x^2-3\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
e, \(6x^2+7xy+2y^2=6x^2+3xy+4xy+2y^2=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)
phan tich da thuc thanh nhan tu 6x4+x3-3x2-17x-5
phan tich da thuc thanh nhan tu : a) 3x^2 - 22xy + 4x + 8y + 7x^2 + 1 ; b) 12x^2 + 5x - 12y^2 + 12y - 10xy - 3 ; c)x^4 + 6x^3 + 11x^2 + 6x + 1
phan tich da thuc thanh nhan tu a, (3x+1)^2-(x+1)^2
b, 6x-6y-x^2+xy
\(a,\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
\(b,6x-6y-x^2+xy\)
\(=\left(6x-6y\right)-\left(x^2-xy\right)\)
\(=6\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left(6-x\right)\)
phan tich da thuc thanh nhan tu (x+1)(x+2)(x+4)(x+5)-40
b)2x3+3x2+6x+5
c)x4-4x3-19x7+106x-120
phan tich da thuc thanh nhan tu (x+1)(x+2)(x+4)(x+5)-40
b)2x3+3x2+6x+5
c)x4-4x3-19x7+106x-120
phan tich da thuc thanh nhan tu
a) x^3+x+2
b) x^3+3x^2-4
c) x^4+x^3+6x^2+5x+5
a)\(3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
b)\(8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
c)\(8x^2-2x-1=8x^2+2x-4x-1=2x\left(4x+1\right)-\left(4x+1\right)=\left(2x-1\right)\left(4x+1\right)\)
phan tich da thuc sau thanh nhan tu
x^3+3x^2+4x+2
6x^4-x^3-7x^2+x+1
giup minh nhe cam on nhieu
\(a.x^3+3x^2+4x+2\)
\(=x^3+x^2+2x^2+2x+2\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+2\right)\)
\(b.6x^4-x^3-7x^2+x+1\)
\(=6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1\)
\(=6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(6x^3+5x^2-2x-1\right)\)
\(=\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)\)
\(=\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[3x\left(2x-1\right)+\left(2x-1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)\)
k giùm cái cho đỡ buồn!