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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
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
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) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)
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, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)
\(\Leftrightarrow\dfrac{xy-12}{4y}=\dfrac{5}{8}\)
=>2(xy-12)=5y
=>2xy-24=5y
=>2xy-5y=24
=>y(2x-5)=24
mà x,y là số nguyên
nên \(\left(2x-5;y\right)\in\left\{\left(1;24\right);\left(-1;-24\right);\left(3;8\right);\left(-3;-8\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;24\right);\left(2;-24\right);\left(4;8\right);\left(1;-8\right)\right\}\)
\(a.\)
\(\dfrac{x}{2}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)
\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)
\(b.\)
\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)
\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)
\(c.\)
\(\dfrac{x}{4}=\dfrac{y}{5}\)
\(\Leftrightarrow\dfrac{2x}{8}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)
mà x+y=35
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)
Vậy: (x,y)=(10;25)
b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)
nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)
hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)
mà y-3x=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
Vậy: (x,y)=(1;5)
c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)
nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)
mà 2x-y=15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
Vậy: (x,y)=(20;25)
Thêm hộ mình ( vô lí ) ktm đề bài vì x;y nguyên nhé