=(x^5-5x^2)-(2xy+10y)

=(x^5-5x^2)-(2xy-10y)

=x^2.(x-5)-2y.(x-5)

=(x^2-2y).(x-5)

k nha

\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)

\(=2xy+5x+4y^2+10y\)

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

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

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

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

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

xong phân tích nốt cái bậc 2 kia để được max điểm :)))

a: x^2+4xy-21y^2

\(=x^2+7xy-3xy-21y^2\)

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

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

b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

=5x(x+y)+y(x+y)

=(x+y)(5x+y)

c: \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

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

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

d: \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

=x(x-2y)-5y(x-2y)

=(x-2y)(x-5y)

a) \(x^2+4xy-21y^2\)

\(=x^2+7xy-3xy-21y^2\)

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

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

b) \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

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

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

c) \(x^2+2xy-15y^2\)

\(=x^2+5xy-3xy-15y^2\)

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

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

d) \(x^2-7xy+10y^2\)

\(=x^2-2xy-5xy+10y^2\)

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

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

a) x² + 4xy - 21y²

= x² - 3xy + 7xy - 21y²

= (x² - 3xy) + (7xy - 21y²)

= x(x - 3y) + 7y(x - 3y)

= (x - 3y)(x + 7y)

b) 5x² + 6xy + y²

= 5x² + 5xy + xy + y²

= (5x² + 5xy) + (xy + y²)

= 5x(x + y) + y(x + y)

= (x + y)(5x + y)

c) x² + 2xy - 15y²

= x² + 2xy + y² - 16y²

= (x² + 2xy + y²) - 16y²

= (x + y)² - (4y)²

= (x + y - 4y)(x + y + 4y)

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

d) x² - 7xy + 10y²

= x² - 2xy - 5xy + 10y²

= (x² - 2xy) - (5xy + 10y²)

= x(x - 2y) - 5y(x - 2y)

= (x - 2y)(x - 5y)

\(x^2-2xy+5x-10y\)

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

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

\(x^2-2xy+5x-10y\)

\(=\left(x^2-2xy\right)+\left(5x-10y\right)\)

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

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

\(x-3\sqrt{x}+\sqrt{xy}-3y\)

\(=\left(x-3\sqrt{x}\right)+\left(\sqrt{xy}-3y\right)\)

\(=\sqrt{x}\left(\sqrt{x}-3\right)+y\left(\sqrt{x}-3\right)\)

\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}+y\right)\)

 ✰ ღ๖ۣۜDαɾƙ ๖ۣۜBαηɠ ๖ۣۜSĭℓεηтღ✰Giỏi thiệt \(\sqrt{xy}=y\sqrt{x}\)

Đọc lại sách lớp 9, bài LIÊN HỆ GIỮA PHÉP NHÂN VÀ PHÉP KHAI PHƯƠNG

Bạn xem lại đề

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

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

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

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

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

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

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

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

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)