\(a,=\left(x+y\right)\left(y+z\right)\\ b,=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ c,=\left(x-y\right)\left(x+y\right)+\left(x-y\right)=\left(x+y+1\right)\left(x-y\right)\\ d,= \left(2x-5\right)\left(2x+5\right)\\ e,=\left(4y-3\right)\left(4y+3\right)\)

a: x/2=-5/y

=>xy=-10

=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)

b: =>xy=12

mà x>y>0

nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

c: =>(x-1)(y+1)=3

=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)

d: =>y(x+2)=5

=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)

a)(x+1)(y-2)=3

x+1;y-2 thuộc Ư(3){1;-1;3;-3}

ta có bảng sau :

vậy cặp x;y thuộc {(2;3);(0;1);(4;5);(-2;-1)}
 

a) Ta có bảng sau:

b) Ta có bảng sau:

\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)

\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)

⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)

\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)

\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)

\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)

`A)2/3=x/60`

`=>40/60=x/60`

`=>x=40`

`B)-1/2=y/18`

`=>-9/18=y/18`

`=>y=-9`

`C)3/x=y/35=-36/84`

Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`

`=>3/x=-3/7`

`=>x=-7`

`y/35=-3/7=-15/35`

`=>y=-15`

`D)7/x=y/27=-42/54`

Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`

`=>7/x=-7/9`

`=>x=-9`

`y/27=-7/9=-21/27`

`=>y=-21`

a) Ta có: \(\dfrac{2}{3}=\dfrac{x}{60}\)

\(\Leftrightarrow x=\dfrac{2\cdot60}{3}=40\)

Vậy: x=40