a: \(x^2-4xy+4y^2-2x+4y-35\)

\(=\left(x^2-4xy+4y^2\right)-\left(2x-4y\right)-35\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)-35\)

\(=\left(x-2y\right)^2-7\left(x-2y\right)+5\left(x-2y\right)-35\)

\(=\left(x-2y\right)\left(x-2y-7\right)+5\left(x-2y-7\right)\)

\(=\left(x-2y-7\right)\left(x-2y+5\right)\)

c: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)

\(=x^2y^2+a^2b^2+2xyab+a^2y^2-2aybx+b^2x^2\)

\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)

\(=y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)

\(=\left(x^2+a^2\right)\left(y^2+b^2\right)\)

a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)

f) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).

b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2

= (a + 4 – b)(a + 4 + b).

TK
 

a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).

b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2

= (a + 4 – b)(a + 4 + b).

Lời giải:

a. $a^4+a^3+a^2+a=(a^4+a^3)+(a^2+a)$

$=a^3(a+1)+a(a+1)=(a+1)(a^3+a)=a(a+1)(a^2+1)$
b. $3xy^2+5y-3x^2y+(-5x)=(3xy^2-3x^2y)+(5y-5x)$

$=3xy(y-x)+5(y-x)=(y-x)(3xy+5)$

c. $xy-z+y-xz=(xy+y)-(z+xz)=y(x+1)-z(x+1)=(x+1)(y-z)$

d.

$x^2-bx+ax-ab=(a^2+ax)-(bx+ab)=a(a+x)-b(a+x)=(a+x)(a-b)$

a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)

= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2+7x+10x\right)\)

= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)