\(=\left(x^2+y^2-5\right)^2-\left[2\left(xy+2\right)\right]^2\\ =\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\\ =\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]\\ =\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(\left(x^2+y^2-5\right)^2-4\left(x^2y^2+4xy+4\right)\)

\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)

\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)

\(=\left(x^2+y^2-2xy-9\right)\left(x^2+y^2+2xy-1\right)\)

\(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)

\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)

\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)

\(=\left[\left(x-y\right)^2-9\right]\left[\left(x-y\right)^2-1\right]\)

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

2x²y + xy² - xy

= xy(2x + y - 1)

=xy*2x+xy*y-xy*1

=xy(2x+y-1)

                     \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

     ⇔   \(3\left(x^2-2x+1\right)-3x\left(x-5\right)=1\)

     ⇔        \(3x^2-6x+3-3x^2+15x=1\)

     ⇔                                       \(9x+3=1\)

     ⇔                                             \(9x=1-3\)

     ⇔                                             \(9x=-2\)

     ⇔                                              \(x=\dfrac{-2}{9}\)

Bạn dặt 3 làm nhân tử chung là xong, dơn giản thui :)))))

k nha!!!!

\(6a\left(x+y\right)+3a+3b\)

\(=3\left[a\left(x+y\right)+a+b\right]\)

\(=3\left(ax+ay+a+b\right)\)

cái này trong sách í

h: =3(x^2+3x-10)

=3(x^2+5x-2x-10)

=3[x(x+5)-2(x+5)]

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

e: =x^2-2x-4x+8

=x(x-2)-4(x-2)

=(x-2)(x-4)

h: =x^2-3x-4x+12

=x(x-3)-4(x-3)

=(x-3)(x-4)

Lời giải:
b. $3x^2+9x-30=3(x^2+3x-10)$

$=3[(x^2-2x)+(5x-10)]=3[x(x-2)+5(x-2)]=3(x-2)(x+5)$

e. $x^2-6x+8=(x^2-2x)-(4x-8)=x(x-2)-4(x-2)=(x-4)(x-2)$

h. $x^2-7x+12=(x^2-3x)-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)$

ai giải gimf nha

phần này học từ kì 1 rồi mà bn

co cau hoi nao dao ma bao giai

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

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)

\(3xy-21x^2y^2=3xy\left(1-7xy\right)\)

\(3x^2y-5xy^2+7xy\)

\(=xy\left(3x-5y+7\right)\)

\(3xy-21x^2y^2\\ =3xy-3xy.7xy\\ =3xy\left(1-7xy\right)\)

Cái thứ hai không phân tích bằng đặt nhân tử chung được đâu nha=)