phan tich da thuc thanh nhan tu
16x(x+1)-(4x+3)(4x-3)
thanks
phan tich da thuc thanh nhan tu
A=x^6-2x^5-4x^4+6x^3+4x^2-2x-1
X^4+4x^3+5x^2+2x+1
Phan tich da thuc thanh nhan tu
phan tich da thuc thanh nhan tu : 4x^4+4x^3+5x^2 +2x +1
4x^4+4x^3+5^2+2x+1
phan tich da thuc thanh nhan tu
4x^4+4x^3+5^2+2x+1 = (4x^4+4x^3+x^2) + (4x^2+2x) + 1 = x^2(2x+1)^2 + 2x(2x+1) + 1 = [x(2x+1)]^2 +2x(2x+1) + 1 = (2x^2+x+1)^2
nếu là 5^2 thì như tui
còn 5x^2 thì như Kami
phan tich cac da thuc sau thanh nhan tu a)x^2+4x+3 b) 4x^2+4x-3 c) x^2-x-12 d)4x^4+4x^2y^2-8y^4
a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)
b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)
c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)
d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)
a) \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=\left(x^2+x\right)+\left(3x+3\right)\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
c) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
\(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
1)x^3-x^2+4
2)4x^4+4x^3-x^2-x
3)x^4+4a^4
phan tich da thuc thanh nhan tu
\(4x^4+4x^3-x^2-x\)
\(=4x^3.\left(x+1\right)-x.\left(x+1\right)=\left(4x^3-x\right).\left(x+1\right)\)
(x^2+x)^2+4x^2+4x-12 phan tich da thuc thanh nhan tu
Phan tich da thuc thanh nhan tu
\(\left(xy+4\right)^2-4\left(x+y\right)^2\)
\(x^4-4x^3+8x^2-16x+16\)
Phan tich da thuc thanh nhan tu x^2(1-x^2)-4-4x^2