d: \(\frac{x}{6}-\frac{2}{y}=\frac{1}{30}\)

=>\(\frac{x}{6}-\frac{1}{30}=\frac{2}{y}\)

=>\(\frac{5x-1}{30}=\frac{2}{y}\)

=>\(y\left(5x-1\right)=60\)

=>(5x-1;y)∈{(1;60);(60;1);(-1;-60);(-60;-1);(2;30);(30;2);(-2;-30);(-30;-2);(3;20);(20;3);(-3;-20);(-20;-3);(4;15);(15;4);(-4;-15);(-15;-4);(5;12);(12;5);(-5;-12);(-12;-5);(6;10);(10;6);(-6;-10);(-10;-6)}

=>(5x;y)∈{(2;60);(61;1);(0;-60);(-59;-1);(3;30);(31;2);(-1;-30);(-29;-2);(4;20);(21;3);(-2;-20);(-19;-3);(5;15);(16;4);(-3;-15);(-14;-4);(6;12);(13;5);(-4;-12);(-11;-5);(7;10);(11;6);(-5;-10);(-9;-6)}

=>(x;y)∈{(2/5;60);(61/5;1);(0;-60);(-59/5;-1);(3/5;30);(31/5;2);(-1/5;-30);(-29/5;-2);(4/5;20);(21/5;3);(-2/5;-20);(-19/5;-3);(1;15);(16/5;4);(-3/5;-15);(-14/5;-4);(6/5;12);(13/5;5);(-4/5;-12);(-11/5;-5);(7/5;10);(11/5;6);(-1;-10);(-9/5;-6)}

c: \(\frac{5}{x}-\frac{y}{3}=\frac16\)

=>\(\frac{5}{x}=\frac{y}{3}+\frac16=\frac{2y+1}{6}\)

=>x(2y+1)=30

=>(x;2y+1)∈{(1;30);(30;1);(-1;-30);(-30;-1);(2;15);(15;2);(-2;-15);(-15;-2);(3;10);(10;3);(-3;-10);(-10;-3);(5;6);(6;5);(-5;-6);(-6;-5)}

=>(x;2y)∈{(1;29);(30;0);(-1;-31);(-30;-2);(2;14);(15;1);(-2;-16);(-15;-3);(3;9);(10;2);(-3;-11);(-10;-4);(5;5);(6;4);(-5;-7);(-6;-6)}

=>(x;y)∈{(1;29/2);(30;0);(-1;-31/2);(-30;-1);(2;7);(15;1/2);(-2;-8);(-15;-3/2);(3;9/2);(10;1);(-3;-11/2);(-10;-2);(5;5/2);(6;2);(-5;-7/2);(-6;-3)}

b: \(\frac{4}{x}+\frac{y}{3}=\frac56\)

=>\(\frac{4}{x}=\frac56-\frac{y}{3}=\frac{5-2y}{6}\)

=>x(5-2y)=24

=>x(2y-5)=-24

=>(x;2y-5)∈{(1;-24);(-24;1);(-1;24);(24;-1);(2;-12);(-12;2);(-2;12);(12;-2);(3;-8);(-8;3);(-3;8);(8;-3);(4;-6);(-6;4);(-4;6);(6;-4)}

=>(x;2y)∈{(1;-19);(-24;6);(-1;29);(24;4);(2;-7);(-12;7);(-2;17);(12;3);(3;-3);(-8;8);(-3;13);(8;2);(4;-1);(-6;9);(-4;11);(6;1)}

=>(x;y)∈{(1;-19/2);(-24;3);(-1;29/2);(24;2);(2;-7/2);(-12;7/2);(-2;17/2);(12;3/2);(3;-3/2);(-8;4);(-3;13/2);(8;1);(4;-1/2);(-6;9/2);(-4;11/2);(6;1/2)}

a: \(\frac{x}{3}-\frac{4}{y}=\frac15\)

=>\(\frac{x}{3}-\frac15=\frac{4}{y}\)

=>\(\frac{5x-3}{15}=\frac{4}{y}\)

=>y(5x-3)=60

=>(5x-3;y)∈{(1;60);(60;1);(-1;-60);(-60;-1);(2;30);(30;2);(-2;-30);(-30;-2);(3;20);(20;3);(-3;-20);(-20;-3);(4;15);(15;4);(-4;-15);(-15;-4);(5;12);(12;5);(-5;-12);(-12;-5);(6;10);(10;6);(-6;-10);(-10;-6)}

=>(5x;y)∈{(4;60);(63;1);(2;-60);(-57;-1);(5;30);(33;2);(1;-30);(-27;-2);(6;20);(23;3);(0;-20);(-17;-3);(7;15);(18;4);(-1;-15);(-12;-4);(8;12);(15;5);(-2;-12);(-9;-5);(9;10);(13;6);(-3;-10);(-7;-6)}

=>(x;y)∈{(4/5;60);(63/5;1);(2/5;-60);(-57/5;-1);(1;30);(33/5;2);(1/5;-30);(-27/5;-2);(6/5;20);(23/5;3);(0;-20);(-17/5;-3);(7/5;15);(18/5;4);(-1/5;-15);(-12/5;-4);(8/5;12);(3;5);(-2/5;-12);(-9/5;-5);(9/5;10);(13/5;6);(-3/5;-10);(-7/5;-6)}

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

=5^2-2*3

=25-6

=19

x^3+y^3=(x+y)^3-3xy(x+y)

=5^3-3*3*5

=125-9*5

=80

(x-y)^2=(x+y)^2-4xy=5^2-4*3=13

=>\(x-y=\sqrt{13}\)

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

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

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

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

Rxfxtd

Bài 1 : 

\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\Rightarrow x=16;y=24;z=30\)

bài 2 : 

Đặt \(x=2k;y=5k\Rightarrow xy=10k^2=10\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)

Với k = 1 thì x = 2 ; y = 5

Với k = - 1 thì x = -2 ; y = -5

b: 4/x+y/3=5/6

=>\(\dfrac{12+xy}{3x}=\dfrac{5}{6}=\dfrac{5x}{6x}\)

=>24+2xy=5x

=>5x-2xy=24

=>x(5-2y)=24

=>x(2y-5)=-24

=>(x;2y-5) thuộc {(24;-1); (-24;1); (8;-3); (-8;3)}(Vì x và y là số nguyên)

=>(x,y) thuộc {(24;2); (-24;3); (8;1); (-8;1)}

a: =>\(\dfrac{xy-12}{3y}=\dfrac{1}{5}\)

=>5(xy-12)=3y

=>5xy-3y=60

=>y(5x-3)=60

=>(y;5x-3) thuộc {(5;12); (30;2)}(Vì x,y là số nguyên)

=>(y,x) thuộc {(5;3); (30;1)}

b: Bạn ghi lại đề đi bạn

b) 4/x+y/3 = 5/6 .