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
phan tich da thuc thanh nhan tu: 4x4+4x3+5x2+2x+1
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 x^3-5x^2+2x+8
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 da thuc thanh nhan tu
\(2x^3-x^2+5x+3\)
$ 2x^3 - x^2 + 5x + 3 \\ = 2x^3 + x^2 - 2x^2 - x + 6x + 3 \\ = x^2(2x + 1) - x(2x + 1) + 3(2x + 1) \\ = (2x + 1)(x^2 - x + 3) $
\(2x^3-x^2+5x+3\)
= \(2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
Vì \(x^2-x+3=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+3>0\)
Nên
\(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
a, x^3-x^2-4x^2+8x-4
b, 4x^2-25-(2x-5)2x+7
c, x^3+27+(x+3)(x-9)
d, 2x^2-2y^2+5x-5y
e, x^2-y^2-2y-1
phan tich da thuc thanh nhan tu
phan tich da thuc thanh nhan tu a. x^3+x+2
b, x^4+5x^3+10x-4
\(x^3+x+2=\left(x^3+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
\(b,x^4+5x^3+10x-4=\left(x^4-4\right)+\left(5x^3-10x\right)\)\(=\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2+5x\right)\)
phan tich da thuc thanh nhan tu 2x(x + 3) + 2(x + 3)
tìm x
a) 5x(x – 2) – x – 2 = 0
b) 4x(x + 1) = 8( x + 1)
c) x(2x + 1) + 1/2 x 3/3 - = 0
d) x(x – 4) + (x – 4)^2 = 0