\(\dfrac{x}{7}+\dfrac{1}{y}=-\dfrac{1}{14}\Leftrightarrow\dfrac{xy+7}{7y}=\dfrac{\dfrac{-y}{2}}{7y}\\ \Leftrightarrow xy+7=-\dfrac{y}{2}\\ 2xy+14=-y\\ y\left(2x+1\right)=-14\)

Vì y,x là số nguyên nên 2x-1 là ước lẻ của -14 = {1;-1;7;-7}

Ta có bảng sau:

Vậy (x,y) thuộc {(0,-14);(-1,14);(3,-2);(-4,2)}

vậy x và y e (-1,14),(0,-14),(3,-2),(-4,2)

Vì x/7+1/y=-1/14 

=xy+7/7y=2/7y

xy+7=y/-2                  (y/-2=-y/2)

2yx+14=-y

y.(2x+1)=-14

X và Y là số nguyên 

2x-1 ước số lẻ của -14 :-7,-1,1,7

X =0,-1,3,-4

Y=-14,-2,2,14

C

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

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

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

Bài 4:

a) \(\dfrac{2.7.13}{26.35}=\dfrac{2.7.13}{13.2.7.5}=\dfrac{1}{5}\)

b) \(\dfrac{23.5-23}{4-27}=\dfrac{23.\left(5-1\right)}{-23}=\dfrac{23.4}{-23}=-4\)

c) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)

Bài 1

a) \(\dfrac{x}{15}=\dfrac{17}{51}\)

51x=17.15

51x=255

⇒x=5

b) \(\dfrac{-7}{y}=\dfrac{12}{24}\)

-7.24=24y

-168=12y

⇒y=-14

;-; giúp em với cần gấp ạ:vvvv

\(\Rightarrow\left(x-1\right).\left(x-1\right)=-4.\left(-4\right)\)

\(\Rightarrow\left(x-1\right)^2=16\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: ...

tham khảo:
(x-1).(x-1)=-4.(-4)
(x-1).(x-1)=16
x.(1.1)=16
x.0=16
x=16:0
x=0

`1/x-1/y=2`
`đk:x,y ne 0`
Nhân 2 vế với `xy ne 0` ta có:
`y-x=2xy`
`=>2y-2x=4xy`
`=>4xy-2x=2y`
`=>2x(2y-1)=2y-1+1`
`=>(2y-1)(2x-1)=1`
Vì `x,y ne 0=>2x-1,2y-1 ne -1`
`=>2x-1=2y-1=1`
`=>x=y=1`

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow\dfrac{x-y}{xy}=\dfrac{1}{2}\)

do x,y nguyên nên:\(\left\{{}\begin{matrix}x-1=1\\x.y=2\end{matrix}\right.\)

ta có :2=1.2

mã-y=1⇒x>y⇒x=2,y=1

Giải:

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{2}\) 

\(\Rightarrow\dfrac{\left(y-x\right)}{\left(xy\right)}=\dfrac{1}{2}\) 

\(\Rightarrow2.\left(y-x\right)=xy\) 

\(\Rightarrow2y-xy-2x=0\) 

\(\Rightarrow y.\left(2-x\right)+2.\left(2-x\right)=4\) 

\(\Rightarrow\left(y+2\right).\left(2-x\right)=4\) 

\(\Rightarrow\left(y+2\right)\) và \(\left(2-x\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)  

Ta có bảng giá trị:

Vậy \(\left(x;y\right)=\left\{\left(3;-6\right);\left(4;-4\right);\left(6;-3\right);\left(-2;-1\right);\left(0;0\right);\left(1;2\right)\right\}\)