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a) \(4x^3y-12x^2y^3-8x^4y^3\)
\(=4x^2y\left(x-3y^2-2x^2y^2\right)\)
b) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
c) \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-y-1\right)\left(x+y-1\right)\)
d) \(x\left(x-2y\right)+3\left(2y-x\right)\)
\(=x\left(x-2y\right)-3\left(x-2y\right)\)
\(=\left(x-3\right)\left(x-2y\right)\)
e) \(x^2+4\)
\(=\left(x^4+4x^2+4\right)-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
f) \(5x^2-7x-6\)
\(=\left(5x^2-10x\right)+\left(3x-6\right)\)
\(=5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(5x+3\right)\left(x-2\right)\)
1 ) x3 - 2x2 + x
= x( x2 - 2x + 1 )
= x ( x-1)2
2) 4x3 - 25x
= x ( 4x2 - 25)
= x( 2x-5) ( 2x +5)
11) \(x^2-y^2-4x+4\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
13) \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
1) \(x^6-x^4-9x^3+9x^2\)
\(=x^2\left(x^4-x^2-9x+9\right)\)
\(=x^2\left[x^2\left(x^2-1\right)-9\left(x-1\right)\right]\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
2) \(x^4-4x^3+8x^2-16x+16\)
\(=x^2\left(x^2+4\right)-4x\left(x^2+4\right)+4\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
3) \(x^4-25x^2+20x-4=x^4+5x^3-2x^2-5x^3-25x^2+10x+2x^2+10x-4\)
\(=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
4) \(5x\left(x-2y\right)+2\left(2y-x\right)^2\)\(=5x\left(x-2y\right)+2\left(x-2y\right)^2=\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
5) \(x^2\left(x^2-6\right)-x^2+9=x^4-7x^2+9\)
\(=x^4+x^3-3x^2-x^3-x^2+3x-3x^2-3x+9\)
\(=x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-3\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x^2-x-3\right)\)
6) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)^2+\left(y-4\right)^3=\left(y-4\right)^2\left(7x+y-4\right)\)
7) \(x^3+2x^2-6x-27=x^3-3x^2+5x^2-15x+9x-27\)
\(=x^2\left(x-3\right)+5x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2+5x+9\right)\)
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)