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a) x4 - x3y - x + y
= x3(x - y) - (x - y)
=(x - y)(x3 - 1)
=(x - y)(x - 1)(x2 + x + 1)
b) x3 - 4x2 - 8x + 8
= (x3 + 8) - (4x2 + 8x)
= (x + 2)(x2 - 2x + 4) - 4x(x + 2)
= (x + 2)(x2 - 2x + 4 - 4x)
= (x + 2)(x2 - 6x + 4)
a, = (x^4-x^3y)-(x-y)
=x^3.(x-y)-(x-y) = (x-y).(x^3-1) = (x-y).(x-1).(x^2+x+1)
b, = (x^3+2x^2)-(6x^2+12x)+(4x+8)
= x^2.(x+2)-6x.(x+2)+4.(x+2) = (x+2).(x^2-6x+4)
k mk nha

\(3x^2-8x+4\)
\(=3x^2-6x-2x+4\)
\(=\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(3x-2\right)\left(x-2\right)\)
a) \(3x^2-8x-4\)
\(=3x^2-6x-2x+4\)
\(=3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
b) \(4x^4+81\)
\(=x^4+81+18x^2-18x^2\)
\(=\left[\left(x^2\right)^2+2x^2.9+9^2\right]-18x^2\)
\(=\left(x^2+9\right)^2-(\sqrt{18}x^2)\)
\(=\left(x^2+9-\sqrt{18}x\right)\left(x^2+9+\sqrt{18}x\right)\)

a) \(x^3-5x^2+8x-4=\left(x^3-x^2\right)-4\left(x^2-x\right)+4\left(x-1\right)=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\) chia hết cho 2x-3 => 7 chia hết cho 2x -3
=> 2x -3 thuộc U(7) ={-7;-1;1;7}
+2x-3 =-7 => x =-2
+2x-3 =-1 => x =1
+2x-3 =1 => x =2
+2x -3 =7 => x =5

( x2 + 8x - 34)2 - ( 3x2 - 8x + 2)2
=> x4 + 8x2 - 342 - 3x4 + 8x2 - 22
=> x4( 1 - 4 ) + ( 8x2 - 8x2 ) + ( 342 - 22 )
-3x4 + 0 + 1152
= -3x4 + 1152
Lk t nhá, Michiel na`h, h pải dùng nik ko có tn

a)\(2a^3+16=2\left(a^3+8\right)=2\left(a+2\right)\left(a^2-2a+4\right)\)
b)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+\left(3y\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2\)
\(=\left(2x+3y\right)^2\)
c)\(x^4-2x^3-x^2+2x+1=\left(x^4-x^3-x^2\right)-\left(x^3-x^2-x\right)-\left(x^2-x-1\right)\)
\(=x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)-\left(x^2-x-1\right)\)
\(=\left(x^2-x-1\right)\left(x^2-x-1\right)=\left(x^2-x-1\right)^2\)

\(x^4-4x^3-8x^2+8x=x\left(x^3-4x^2-8x+8\right)=x\left[\left(x^3+8\right)-\left(4x^2+8x\right)\right]=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)=x\left(x+2\right)\left(x^2-6x+4\right)\)


\(\left(x-5\right)^2-16=\left(x-5\right)^2-4^2=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)
\(25-\left(3-x\right)^2=5^2-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
\(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)\(49\left(y-4\right)^2-9\left(y+2\right)^2=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left(7y-28-3y-6\right)\left(7y-28+3y-6\right)=\left(4y-34\right)\left(10y-22\right)\)\(=4.\left(2y-17\right)\left(5y-11\right)\)
e); f) Áp dụng hằng đẳng thức số 6,7 để làm
4x2-8x-5
=4x2-10x+2x-5
=2x(2x-5)+1(2x-5)
=(2x-5)(2x+1)