\(\Leftrightarrow\dfrac{xy-12}{4y}=\dfrac{5}{8}\)

=>2(xy-12)=5y

=>2xy-24=5y

=>2xy-5y=24

=>y(2x-5)=24

mà x,y là số nguyên

nên \(\left(2x-5;y\right)\in\left\{\left(1;24\right);\left(-1;-24\right);\left(3;8\right);\left(-3;-8\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;24\right);\left(2;-24\right);\left(4;8\right);\left(1;-8\right)\right\}\)

y = 8

x = 4

Giải:

a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\) 

\(\Rightarrow x\in\left\{0;\pm1;2\right\}\) 

b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow-12.\left(x-6\right)=4.18\) 

\(\Rightarrow-12x+72=72\) 

\(\Rightarrow-12x=72-72\) 

\(\Rightarrow-12x=0\) 

\(\Rightarrow x=0:-12\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\) 

\(\dfrac{x+46}{20}=\dfrac{2x}{5}\) 

\(\Rightarrow5.\left(x+46\right)=2x.20\) 

\(\Rightarrow5x+230=40x\) 

\(\Rightarrow5x-40x=-230\) 

\(\Rightarrow-35x=-230\) 

\(\Rightarrow x=-230:-35\) 

\(\Rightarrow x=\dfrac{46}{7}\) 

Chúc bạn học tốt!

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

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

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

tìm các số nguyên x , y biết  5\x + 4\y = 1\8 - Hoc24

trên mạng hônk có :v

Lời giải:

a. $\frac{x}{7}=\frac{6}{21}$

$x=\frac{6}{21}.7$

$x=2$

b.

$\frac{-5}{y}=\frac{20}{28}$

$y=-5:\frac{20}{28}$

$y=-7$

c.

$\frac{-4}{8}=\frac{-7}{y}$

$y=-7:\frac{-4}{8}$

$y=14$

a, \(\dfrac{x}{7}=\dfrac{6}{21}\Leftrightarrow\dfrac{3x}{21}=\dfrac{6}{21}\Rightarrow x=2\)

b, \(\dfrac{-5}{y}=\dfrac{20}{28}\Leftrightarrow\dfrac{20}{-4y}=\dfrac{20}{28}\Leftrightarrow y=-7\)

c, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)

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

b) \(Q=\dfrac{27-2x}{12-x}=\dfrac{2.\left(12-x\right)+3}{12-x}=2+\dfrac{3}{12-x}\)

Để Q đạt max 

thì \(\dfrac{3}{12-x}\) phải max nên 12 - x phải min và 12 - x > 0 

lại có \(x\inℤ\) 

nên 12 - x = 1 

<=> x = 11 

Khi đó Q = 17

Vậy Q_max = 5 khi x = 11 

x = 6; y=2