a: 2(x+y)=5(y+z)=3(x+z)

=>\(\frac{2\left(x+y\right)}{30}=\frac{5\left(y+z\right)}{30}=\frac{3\left(x+z\right)}{30}\)

=>\(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}\)

Đặt \(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}=k\)

=>x+y=15k; y+z=6k; x+z=10k

=>x+y-x-z=15k-10k=5k; y+z=6k

=>y-z=5k và y+z=6k

=>y=(5k+6k)/2=5,5k; z=5,5k-5k=0,5k

x+y=15k

=>x+5,5k=15k

=>x=9,5k

x-y=9,5k-5,5k=4k; y-z=5,5k-0,5k=5k

=>\(\frac{x-y}{4}=\frac{y-z}{5}\)

a: 5x+y⋮19

=>3(5x+y)⋮19

=>19x-3(5x+y)⋮19

=>19x-15x-3y⋮19

=>4x-3y⋮19

b: 7x+2y=39x+26y-32x-24y

=13(3x+2y)-8(4x+3y)

mà 13(3x+2y)⋮13 và 8(4x+3y)⋮13(Do 4x+3y⋮13)

nên 7x+2y⋮13

\(a,\left(x+y+z\right)^3-x^3-y^3-z^3\\ =\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\\ =\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\\ =x^3+y^3+z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\\ =\left(x+y\right)\left(3xy+3xz+3yz+3z^2\right)\\ =3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\\ =3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(b,x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy-3xy\right)\\ =0\left(x^2+y^2+z^2-xz-yz-xy\right)=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)

ta có 4x - 3y = 19x - 3.(5x + y) 

Vì 19x chia hết cho 19;

5x + y chia hết cho 19 nên 3(5x + y) chia hết cho 19

do đó 19x - 3(5x + y) chia hết cho 19 hay 4x - 3y chia hết cho 19

vì 5x+y : 19 nên

5x:19 =>x:19=>4x:19(1)

y:19 =>3y:19 (2)

từ 1 và 2 ta có

4x-3y:19

(dấu : là chia hết)