3x=4y

nên x/4=y/3

Đặt x/4=y/3=k

=>x=4k; y=3k

\(H=\dfrac{2xy+3x^2}{3xy+4y^2}=\dfrac{2\cdot4k\cdot3k+3\cdot16k^2}{3\cdot4k\cdot3k+4\cdot9k^2}\)

\(=\dfrac{24k^2+48k^2}{36k^2+36k^2}=1\)

\(a,3x\left(3x+6\right)=9x^2+18x\)

\(b,-\dfrac{1}{2}xy\left(4x^2+6x\right)\)

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

\(c,-2x^2y^3\left(\dfrac{1}{2}xy+4y^2\right)\)

\(=-x^3y^4-8x^2y^5\)

\(d,-6x^2\left(\dfrac{1}{3}xy^2-\dfrac{1}{2}y\right)\)

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

#\(Urushi\)

a) 3x(3x+6) = 9x^2 + 18x b) -1/2xy(4x^2+6x) = -2x^3y - 3xy c) -2x^2y^3(1/2xy+4y^2 ) = -x^2y^2 - 8x^2y^5 d) -6x^2(1/3xy^2-1/2y) = -2xy + 3x^2y

Vậy kết quả của các biểu thức là: a) 9x^2 + 18x b) -2x^3y - 3xy c) -x^2y^2 - 8x^2y^5 d) -2xy + 3x^2y

a: M=2(-2x-3xy^2+1)-3xy^2+1

=-4x-6xy^2+2-3xy^2+1

=-4x-9xy^2+3

b: Thay x=-2 và y=3 vào M, ta được:

M=2*(-2)-3*(-2)*3^2+1

=-4+1+6*9

=54-3

=51

a. \(2x^2y^3.\frac{1}{4}xy.\left(-3xy\right)=-\frac{3}{2}x^4y^5\text{ đa thức có bậc 4+5 = 9}\)

b. \(\left(-3xy^3\right)^3\left(-\frac{2}{3}x^4y\right)=-27x^3y^9\left(-\frac{2}{3}x^4y\right)=18x^7y^{10}\text{ có bậc 7+10 = 17}\)

c.. \(\frac{2}{3}xy^2-2xy+4x^2y+12+2xy^2-3xy-20-4x^2y=\frac{8}{3}xy^2-5xy-8\) có bậc 3

a) \(M-\left(x^2y-1\right)=-2x^3+x^2y+1\)

\(\Rightarrow M-x^2y+1=-2x^3+x^2y+1\)

\(\Rightarrow M=-2x^3+x^2y+1+x^2y-1\)

\(\Rightarrow M=-2x^3+2x^2y\)

b) \(3x^2+3xy-x^3-M=3x^2+2xy-4y^2\)

\(\Rightarrow-M=3x^2+2xy-4y^2-3x^2-3xy+x^3\)

\(\Rightarrow-M=x^3-4y^2-xy\)

\(\Rightarrow M=-x^3+4y^2+xy\)

a: \(A=-18x^3y^4z\)

Bậc là 8

b: \(M=3x^2+3xy-x^3-3x^2-2xy+4y^2=-x^3+xy+4y^2\)

a) \(3x^2-3xy-5x+5y\)

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

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

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

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

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

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

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

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

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

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

b,2(xy+xyz-2x-2z)

c, 3(x^2-xy-x-y)

a) Ta có : 2x² - 4x + 2xy - 4y 

= 2x(x - 2) + 2y(x - 2) 

= (x - 2)(2x + 2y) 

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