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, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)

=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)

=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}

vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3

xét bảng

vậy giá trị x,y cần tìm là: {x=18.y=2}

{x=-18.y=3}

{x=6, y=1}Ư

{x=-6,y=4}

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

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

các ý a,b,c c chỉ cần nhan chéo cho nhau

Giải:

a) \(\dfrac{12}{16}=\dfrac{-x}{4}=\dfrac{21}{y}=\dfrac{z}{80}\)  

\(\Rightarrow x=\dfrac{12.-4}{16}=-3\) 

\(\Rightarrow y=\dfrac{16.21}{12}=28\) 

\(\Rightarrow z=\dfrac{12.80}{16}=60\) 

b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)\)  =0

    \(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\) 

     \(x.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)\)   \(=0+\dfrac{2}{5}\) 

            \(x.\dfrac{11}{15}\)       \(=\dfrac{2}{5}\) 

                 x          \(=\dfrac{2}{5}:\dfrac{11}{15}\) 

                x           \(=\dfrac{6}{11}\) 

c) (2x-3)(6-2x)=0

⇒2x-3=0 hoặc 6-2x=0

        x=3/2 hoặc x=3

d) \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)

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

               \(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}\)  

                   \(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}\) 

                   \(2x-5=\dfrac{-13}{2}\) 

                         \(2x=\dfrac{-13}{2}+5\)

                         \(2x=\dfrac{-3}{2}\) 

                           \(x=\dfrac{-3}{2}:2\) 

                           \(x=\dfrac{-3}{4}\) 

e) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}\) 

       \(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}:2\) 

       \(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\) 

\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{8}\)  hoặc \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-1}{8}\) 

                \(x=\dfrac{11}{12}\) hoặc \(x=\dfrac{5}{12}\)

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

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

Bài 2:

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

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

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

Lời giải:

$\frac{2x}{3}-\frac{2}{y}=\frac{1}{3}$

$\frac{2xy-6}{3y}=\frac{1}{3}$

$\frac{2xy-6}{3y}=\frac{y}{3y}$

$\Rightarrow 2xy-6=y$

$\Rightarrow y(2x-1)=6$

$y=\frac{6}{2x-1}$
Vì $y$ nguyên nên $\frac{6}{2x-1}$ phải nguyên

$\Rightarrow 2x-1\in Ư(6)$

Mà $2x-1$ lẻ với mọi $x$ nguyên nên $2x-1\in\left\{1;-1;3;-3\right\}$

$\Rightarrow x\in\left\{1; 0; 2; -1\right\}$

Với $x=1$ thì $y=\frac{6}{2x-1}=6$

Với $x=0$ thì $y=\frac{6}{2x-1}=-6$

Với $x=2$ thì $y=\frac{6}{2x-1}=2$

Với $x=-1$ thì $y=\frac{6}{2x-1}=-2$

giúp e bài 6 với ạ. E cám ơn

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

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