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\(x^4+2008x^2+2007x+2008\\ =x^4-x+2008\left(x^2+x+1\right)=x\left(x^3-1\right)+2008\left(x^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Ta có: \(x^4+2008x^2+2007x+2008\)
\(=x^4-x+2008\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)
Do câu d mình ko biết làm bởi v mình không làm được
\(a,=x\left(x^2-4x+4-z^2\right)=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-z-2\right)\left(x+z-2\right)\\ b,=\left(x-y\right)^2-\left(z-5\right)^2=\left(x-y-z+5\right)\left(x-y+z-5\right)\)
\(x^3-4x^2+4x-xz^2=x\left(x^2-4x+4-z^2\right)\)
\(=x\left[\left(x-2\right)^2-z^2\right]=x\left(x-2-z\right)\left(x-2+z\right)\)
\(x^2-2xy+y^2-z^2+10z-25\)
\(=\left(x-y\right)^2-\left(z-5\right)^2\)
\(=\left(x-y+z-5\right)\left(x-y-z+5\right)\)
a, \(6x^3y^2.\left(2-x\right)+9x^2y^2\left(x-2\right)\)
\(=6x^3y^2.\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=y^2.\left(2-x\right)\left(6x^3-9x^2\right)\)
\(=3x^2y^2.\left(2-x\right)\left(2x-3\right)\)
b. \(x^2-4x+4y-y^2\)
\(=\left(x^2-y^2\right)-\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-4\right)\)
a: =(x-z)(y+8)
b; =x^2-2x-3x+6
=(x-2)(x-3)
c: =x^4+10x^2-x^2-10
=(x^2+10)(x^2-1)
=(x^2+10)(x-1)(x+1)
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-3x+2y\right)\)
\(=0\cdot0\)
\(=0\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)^2\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x+y\right)\left(x-y\right)\)
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
