\(N=\dfrac{2}{5}x^2y+xy^2-3xy+\dfrac{1}{3}xy^2-3xy-\dfrac{1}{2}x^2y\)

\(=\left(\dfrac{2}{5}x^2y-\dfrac{1}{2}x^2y\right)+\left(xy^2+\dfrac{1}{3}xy^2\right)+\left(-3xy-3xy\right)\)

\(=-\dfrac{1}{10}x^2y+\dfrac{4}{3}xy^2-6xy\)

\(=-\dfrac{1}{10}.\left(0,5\right)^2.\left(-1\right)+\dfrac{4}{3}.0,5.\left(-1\right)^2-6.0,5.\left(-1\right)\)

\(=\dfrac{1}{40}+\dfrac{2}{3}+3=\dfrac{443}{120}\)

trong mat phang oxy cho tam giac ABC có C 9-2;-5/3),cos BC=4/5,Mthuoc BC,ME vuong goc AB,MF vuong goc AC,I(7/3;1/3) la trung diem AM.tim toa do A biet ym<0

\(VT=\dfrac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)

a: =-4xyz^2

b: =-9x^2y

c: =16x^2y^2

d: =1/6x^2y^3

e: =13/6x^3y^2

f: =7/12x^4y

a) -xyz² - 3xz.yz

= -xyz² - 3xyz²

= -4xyz²

b) -8x²y - x.(xy)

= -8x²y - x²y

= -9x²y

c) 4xy².x - (-12x²y²)

= 4x²y² + 12x²y²

= 16x²y²

d) 1/2 x²y³ - 1/3 x²y.y²

= 1/2 x²y³ - 1/3 x²y³

= 1/6 x²y³

e) 3xy(x²y) - 5/6 x³y²

= 3x³y² - 5/6 x³y²

= 13/6 x³y²

f) 3/4 x⁴y - 1/6 xy.x³

= 3/4 x⁴y - 1/6 x⁴y

= 7/12 x⁴y

a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)

b: \(=-5y-9+xy\)

a, Thay x = 1/2 ; y = -1/3 ta được 

\(A=\dfrac{3.1}{8}\left(-\dfrac{1}{3}\right)+\dfrac{6.1}{4}.\left(\dfrac{1}{9}\right)+\dfrac{3.1}{2}\left(-\dfrac{1}{3}\right)^3\)

\(=-\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{3}{2\left(-27\right)}=-\dfrac{7}{72}\)

b, Thay x = -1 ; y = 3 ta được 

\(B=9+\left(-1\right).3-1+27=32\)

bạn thay chỗ nào x là \(\dfrac{1}{2}\) còn chỗ nào y là \(\dfrac{-1}{3}\)nhé

còn như là 3\(x^3\)y thì thành là 3.\(x^3\).y nhé

mk lười nên ko giải ra cho bạn được 

a: \(=-2\cdot\dfrac{1}{4}\cdot3\cdot x^2y^2\cdot xy\cdot xy^3=-\dfrac{3}{2}x^4y^6\)

b: \(=4x^6y^2\cdot xy^2\cdot\dfrac{1}{2}y^5=2x^7y^9\)

a) \(=\left(2.\dfrac{1}{4}.\left(-3\right)\right).\left(x^2.x.x\right).\left(y^2.y^3.y\right)\)

\(=-\dfrac{3}{2}x^4y^6\)

b) \(-4x^6y^2.x.y^2.\dfrac{1}{2}y^5\)

\(=\left(4.\dfrac{1}{2}\right).\left(x^6.x\right).\left(y^2.y^2.y^5\right)\)

\(=2x^7y^9\)

\(a,VT=\dfrac{x^2+2xy+4-3x^2-3xy}{\left(x+y\right)\left(x+2y\right)}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x-2y\right)}=VP\\ b,VP=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}=VT\)

\(\frac{x^2-x-6}{x^2+7x+10}\)

\(=\frac{x^2-3x+2x-6}{x^2+5x+2x+10}=\frac{x.\left(x-3\right)+2.\left(x-3\right)}{x.\left(x+5\right)+2.\left(x+5\right)}\)

\(=\frac{\left(x+2\right).\left(x-3\right)}{\left(x+2\right).\left(x+5\right)}=\frac{x-3}{x+5}\)