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c, x4+6x3+11x2+6x+1
=x4+6x3+9x2+2x2+6x+1
=x4+9x2+1+6x3+2x2+6x
=(x2)2+(3x)2+12+2.x2.3x+2.x2.1+2.3x.1 (1)
Áp dụng hằng đẳng thức (a+b+c)2=a2+b2+c2+2ab+2ac+2bc
=> (1)=(x2+3x+1)2
Câu a nhé bạn:
a, 3x2−22xy−4x+8y+7y2+1
=3x2-21xy-xy-3x-x+7y+y+7y2+1
=(3x2−21xy−3x)−(xy-7y2-y)−(x-7y-1)
=3x(x−7y−1)−y(x−7y−1)−(x−7y−1)
=(3x−y−1)(x−7y−1)
\(A=3x^2-22xy-4x+8y+7y^2+1\)
Giả sử:
\(A=\left(3x+ay+b\right)\left(x+cy+d\right)\)
\(=3x^2+3cxy+3dx+axy+acy^2+ady+bx+bcy+bd\)
\(=3x^2+acy^2+\left(3c+a\right)xy+\left(3d+b\right)x+\left(ad+bc\right)y+bd\)
Ta có:
\(\begin{cases}\begin{matrix}ac=-7\\3c+a=-22\\3d+b=-4\\ad+bc=8\end{matrix}\\bd=1\end{cases}\)\(\Rightarrow\begin{cases}a=-1\\b=-1\\c=-7\\d=-1\end{cases}\)
Vậy \(A=\left(3x-y-1\right)\left(x-7y-1\right)\)
Chúc bạn học tốt ^^
a) \(4x^4+4x^3+5x^2+2x+1=\left[\left(2x^2\right)^2+4x^3+x^2\right]+2\left(2x^2+x\right)+1=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1=\left(2x^2+x+1\right)^2\)
b) \(3x^2+22xy+11x+37y+7y^2+10=\left(3x^2+21xy+6x\right)+\left(7y^2+xy+2y\right)+\left(5x+35y+10\right)\)
\(=3x\left(x+7y+2\right)+y\left(x+7y+2\right)+5\left(x+7y+2\right)\)
\(=\left(3x+y+5\right)\left(x+7y+2\right)\)
c) Không phân tích được.
d) \(x^4-8x+63=\left(x^4+4x^3+9x^2\right)-\left(4x^3+16x^2+36x\right)+\left(7x^2+28x+63\right)\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
c) \(x^4-7x^3+14x^2-7x+1=\left(x^4-3x^3+x^2\right)-\left(4x^3-12x^2+4x\right)+\left(x^2-3x+1\right)\)
\(=x^2\left(x^2-3x+1\right)-4x\left(x^2-3x+1\right)+\left(x^2-3x+1\right)\)
\(=\left(x^2-3x+1\right)\left(x^2-4x+1\right)\)
c) 2x2 + 10x + 8
= 2x2 + 2x + 8x + 8
= 2x( x + 1) + 8(x + 1)
= 2(x + 1)(x + 4)
d) x2 - 7xy + 10y2
= x2 - 2xy - 5xy + 10y2
= x(x - 2y) - 5y(x - 2y)
= (x - 2y)(x - 5y)
e) x4 + 4x2 - 5
= x4 - x2 + 5x2 - 5
= x2(x2 - 1) + 5(x2 - 1)
= (x - 1)(x + 1)(x2 + 5)
f) x3 - 7x - 6
= x3 - x - 6x - 6
= x(x2 - 1) - 6(x + 1)
= x(x - 1)(x + 1) - 6(x + 1)
= (x + 1)(x - 1)(x - 6)
pn coi kt lại nhé