a. 

$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$

b.

$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$

c.

$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$

d.

$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$

$=(x+1)(x^2-4x+1)$

e.

$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$

$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$

f.

$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$

$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$

g.

$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$

$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$

$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$

$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$

h.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$

e) Ta có: \(x^4-2x^3+2x-1\)

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

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

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

\(=\left(x+1\right)\cdot\left(x-1\right)^3\)

h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)

\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)

\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)

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

a) Ta có: \(x^2-y^2-2x-2y\)

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

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

b) Ta có: \(x^2\left(x+2y\right)-x-2y\)

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

\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)

c) Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2-9\right)\)

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

d) Ta có: \(x^4+2x^3+2x-1\)

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

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

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

\(1,=6xy\left(x^2-2xy+y^2\right)=6xy\left(x-y\right)^2\\ 2,=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\\ 3,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ 4,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 5,=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\\ 6,Sửa:x^2-x-2=x^2+x-2x-2=\left(x+1\right)\left(x-2\right)\\ 7,=x^4-4x^2-x^2+4=\left(x^2-4\right)\left(x^2-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\\ 8,=-x^3-x^2-x=-x\left(x^2+x+1\right)\\ 9,=\left(a-3\right)\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\\ =\left(a-3\right)\left(a^2+9a+18\right)\\ =\left(a-3\right)\left(a^2+3a+6a+18\right)\\ =\left(a-3\right)\left(a+3\right)\left(a+6\right)\)

\(10,=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\\ =xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\\ =\left(x-y\right)\left(xy-xz-yz+z^2\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)

a) 3a +3b -a²-ab

= 3.(a+b) -a.(a+b)=(3-a).(a+b)

b) x² +x +y²-y-2xy

=(x² - 2xy+y²) +(x-y)

=(x-y).(x-y+1)

c) -x² +7x -6

= -x² + x +6x-6

= x.(1-x) -6.(1-x) = (1-x).(x-6)

d) 5x³y -10x²y² +5xy³

= 5xy.(x² -2xy +y²) = 5xy.(x-y)²

e) 2x² +7x -15

= 2x² -3x +10x -15

=x.(2x-3) + 5.(2x-3)

=(2x-3).(x+5)

g) x² -2x +2y -xy

=x.(x-2)-y.(x-2)

=(x-y).(x-2)

h) bn go lai de ho mk dc k?

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)

\(=\left(2x-1\right)\left(-3x+2\right)\)

1.

a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

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

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

casio fx 570 thì ấn mode => 5 => 3 sau điền hệ số a;b;c

casio fx 580 thì ấn mode => 9 => 2 => 2 => điền hệ số a;b;c

có cả cách này à =)))

menu setup -> 9 -> 2 - > 2 (pt cần phân tích)  -> nhập hệ số của pt vào từng biến thích hợp -> ''='' 

VD : \(A=x^2+4x-5\)có nghiệm \(x_1=1;x_2=-5\)

vậy đa thức cần phân tích là : \(\left(x-1\right)\left(x+5\right)=x^2+5x-x-5\)

Vậy \(A=x^2+4x-5=x^2+5x-x-5=\left(x-1\right)\left(x+5\right)\)

tương tự nhé 

cùng vị trí nên tao lộn thôi =))

a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)

b: \(=25-\left(x-2y\right)^2\)

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

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

\(=\left(x-y\right)^2+2.\left(x-y\right).2+2^2\)

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

hk tốt

^^