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\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
\(x^6-x^4-9x^3+9x^2\)
\(=x^2\left(x^4-x^2\right)-x^2\left(9x-9\right)\)
\(=x^2\left(x^4-x^2-9x+9\right)\)
\(=x^2\left(x^4+x^3-9x-x^3-x^2+9\right)\)
\(=x^2\left[x\left(x^3+x^2-9\right)-\left(x^3+x^2-9\right)\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
\(x^4-4x^3+8x^2-16x+16\)
\(=x^4-4x^3+4x^2+4x^2-16x+16\)
\(=x^2\left(x^2-4x+4\right)+4\left(x^2-4x+4\right)\)
\(=\left(x^2-4x+4\right)\left(x^2+4\right)\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
1. a) \(8x^3-32x=8x\left(x^2-4\right)=8x\left(x-4\right)\left(x+4\right)\)
b) \(y^3+64+\left(y+4\right)\left(y-16\right)=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)
\(=\left(y+4\right)\left(y^2-4y+16\right)+\left(y+4\right)\left(y-16\right)=\left(y+4\right)\left(y^2-4y+16+y-16\right)\)
\(=\left(y-4\right)\left(y^2-3y\right)=\left(y-4\right)y\left(y-3\right)\)
2) a)
\(4x^3-9x=0\)
\(\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow x\left(2x+3\right)\left(2x-3\right)=0\)
<=> x=0 hoặc 2x+3=0 hoặc 2x-3=0
<=> x=0 hoặc x=-3/2 hoặc x=3/2
b) \(A=x^3-9x^2+27x-27=x^3-3.x^2.3+3.x.3^2-3^3=\left(x-3\right)^3\)
Tại x=203
A=(203-3)3=2003
Bài 1 :
a) \(8x^3-32x\)
\(=8x\left(x^2-4\right)\)
\(=8x\left(x-2\right)\left(x+2\right)\)
b) \(y^3+64+\left(y+4\right)\left(y-16\right)\)
\(=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)
\(=\left(y+4\right)\left(y^2-4y+16\right)+\left(y+4\right)\left(y-16\right)\)
\(=\left(y+4\right)\left(y^2-4x+16+y-16\right)\)
\(=\left(y+4\right)\left(y^2+y-4x\right)\)
Bài 2 :
a) \(4x^3-9x=0\)
\(x\left(4x^2-9\right)=0\)
\(x\left[\left(2x\right)^2-3^2\right]=0\)
\(x\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\2x-3=0\\2x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\\x=\frac{-3}{2}\end{cases}}}\)
P.s: ở trên dùng ngoặc vuông nhé
b) \(A=x^3-9x^2+27x-27\)
\(A=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(A=\left(x-3\right)^3\)
Thay x = 203 vào biểu thức ta có :
\(A=\left(203-3\right)^3\)
\(A=200^3\)
\(A=8000000\)
a) \(4x^4+4x^3-x^2-x=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(4x^3-x\right)\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\)
\(=x\left\{\left(2x\right)^2-1\right\}\left(x+1\right)=x\left(2x-1\right)\left(2x+1\right) \left(x+1\right)\)
c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)=\left(x^2-4x+4\right)\left(x^2+4\right)=\left(x-2\right)^2\left(x^2+4\right)\)
b) \(x^6-x^4-9x^3+9x^2=x^4\left(x^2-1\right)-\left(9x^3-9x^2\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=\left(x^5+x^4-9x^2\right)\left(x-1\right)=\left(x-1\right)x^2\left(x^3+x^2-9\right)\)