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\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\Rightarrow\dfrac{6}{6x}+\dfrac{2xy}{6x}=\dfrac{5x}{6x}\Rightarrow6+2xy=5x\)
\(\Rightarrow5x-2xy=6\Rightarrow x\left(5-2y\right)=6\)
Do \(x,y\) là số tự nhiên nên \(x\inƯ^+\left(6\right)\)
TH1: \(x=1\Rightarrow5-2y=6\Rightarrow y=-\dfrac{1}{2}\) (loại)
TH2: \(x=2\Rightarrow5-2y=3\Rightarrow y=1\) (TM)
TH3: \(x=3\Rightarrow5-2y=2\Rightarrow y=\dfrac{3}{2}\) (Loại)
TH4: \(x=6\Rightarrow5-2y=1\Rightarrow y=2\) (TM)
\(\Leftrightarrow6+2xy=5x\left(x\ne0\right)\)
\(\Leftrightarrow5x-2xy=6\Leftrightarrow x\left(5-2y\right)=6\)
\(\Leftrightarrow x=\dfrac{6}{5-2y}\)
Để x nguyên thì 5-2y phải là ước của 6
\(\Rightarrow5-2y=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow y=\left\{4;3;2;1\right\}\Rightarrow x=\left\{-2;-6;6;2\right\}\)
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, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)
a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
| x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
| y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 1 | 1 | -1 | 3 | -3 |
| y + 1 | 3 | -3 | 1 | -1 |
| x | 2 | 0 | 4 | -2 |
| y | 2 | -4 | 0 | -2 |
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
a: \(\Leftrightarrow-4< =x< =-3\)
hay \(x\in\varnothing\)
b: =>-9<x<=3
hay \(x\in\left\{0;1;2;3\right\}\)
a) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)
\(\dfrac{4}{y}\) = \(\dfrac{x}{3}-\dfrac{1}{5}\)
\(\dfrac{4}{y}\) = \(\dfrac{5x-3}{15}\)
=> 4.15 = y.(5x-3)
60 = y.(5x-3)
Ta có bảng
| 5x-3 | 1 | 60 | 2 | 30 | 3 | 20 | 4 | 15 | 5 | 12 | 6 | 10 |
| y | 60 | 1 | 30 | 2 | 20 | 3 | 15 | 4 | 12 | 5 | 10 | 6 |
| x | 4/5 | 63/5 | 1 | 33/5 | 6/5 | 23/5 | 7/5 | 18/5 | 8/5 | 3 | 9/5 | 13/5 |
| L | L | TM | L | L | L | L | L | L | TM | L | L |
Vậy y=30 và x=1 ; y=5 và x=3
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
=>(12-xy)/3x=5/6
=>6(12-xy)=15x
=>(12-xy)=5/2x
=>24-2xy=5x
=>5x+2xy=24
=>x(2y+5)=24
=>(x;2y+5) thuộc {(1;24); (2;12); (3;8); (4;6); (6;4); (8;3); (12;2); (24;1)}
mà x,y là các số tự nhiên
nên \(\left(x,y\right)\in\varnothing\)