tinh gia tri cua cac da thuc sau biet x+y-2=0.M=x^4+2x^3y-2x^3+x^2y^2-2x^2y-x(x+y)+2x+3
\(\frac{2x}{2}=\frac{3y}{5}khi-đó-\frac{19x}{x+y}\)
a, Tim x biet:/x-2/+/3-2x/=2x+1
b, Tim x,y thuoc Z biet:xy+2x-y=5
c, tim x,y,z, biet :2x=3y;4y=5zva 4x-3y+5z=7
\(tinh\)\(-2x^3y^2+2x^2y^3\)
biet \(xy=3;y-x=1\)
tim (x;y) biet (x+2y -3)^2016 + |2x + 3y - 5| =0
tim x,y,z biet x/4=y/3 . x/3=z/5 và 2x-3y=6
Biết \(\frac{2x}{3}=\frac{3y}{5}.\)Khi đó \(\frac{19x}{x+y}=...?\)
Tim x,y,z biet x/4=y/3 . x/3=z/5 và 2x - 3y + z=6
bt1) TIM X,Y,Z biet:
a) x/y = 3/4 ; y/z = 5/7 va 2x + 3y - z = 186
b) 2x = 3y = 5z va /x+y-z/ = 95