`7/2 : y= 3/8 + 5/4`

`=> 7/2 : y= 3/8 +10/8`

`=> 7/2 : y=13/8`

`=> y= 7/2 : 13/8`

`=> y= 7/2 xx 8/13`

`=> y= 28/13`

\(\dfrac{7}{2}:y=\dfrac{3}{8}+\dfrac{5}{4}\)
\(\dfrac{7}{2}:y=\dfrac{13}{8}\)
\(y=\dfrac{7}{2}:\dfrac{13}{8}\)
\(y=\dfrac{7}{2}\times\dfrac{8}{13}\)
\(y=\dfrac{28}{13}\)

\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)

\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)

a, 2\(\dfrac{2}{5}\): y \(\times\)1\(\dfrac{3}{4}\) = \(\dfrac{7}{8}\)

     \(\dfrac{12}{5}\) : y   \(\times\dfrac{7}{4}\)    = \(\dfrac{7}{8}\)

   \(\dfrac{12}{5}\)  : y         = \(\dfrac{7}{8}\) : \(\dfrac{7}{4}\)

          \(\dfrac{12}{5}\) : y        = \(\dfrac{1}{2}\)       

                  y         = \(\dfrac{12}{5}\) : \(\dfrac{1}{2}\)

                  y         =    \(\dfrac{24}{5}\)

b, 3\(\dfrac{2}{5}\): y : 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\) 

     \(\dfrac{17}{5}\): y: \(\dfrac{5}{4}\)    = \(\dfrac{13}{5}\)

       \(\dfrac{17}{5}\):y         = \(\dfrac{13}{5}\times\dfrac{5}{4}\)

       \(\dfrac{17}{5}\) : y       = \(\dfrac{13}{4}\)

               y        =   \(\dfrac{17}{5}\) : \(\dfrac{13}{4}\)

               y        =  \(\dfrac{68}{65}\)

c, \(\dfrac{12}{5}\) - 2\(\dfrac{2}{5}\)\(\times y\) = 1\(\dfrac{1}{4}\)

     \(\dfrac{12}{5}\) - \(\dfrac{12}{5}\)\(\times\)y = \(\dfrac{5}{4}\)

            \(\dfrac{12}{5}\times y\) =  \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)

           \(\dfrac{12}{5}\) \(\times\) y =    \(\dfrac{23}{20}\)

                     \(y\)  = \(\dfrac{23}{20}\): \(\dfrac{12}{5}\)

                     y   = \(\dfrac{23}{48}\)

Câu đầu em xem lại đề bài sao có hai dấu bằng.

Câu 2: 

\(\dfrac{3}{2}\) \(\times\)y - \(\dfrac{3}{4}\) \(\times\)y + y = \(\dfrac{4}{5}\)

y \(\times\) ( \(\dfrac{3}{2}\) - \(\dfrac{3}{4}\) + 1) = \(\dfrac{4}{5}\)

y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)

y \(\times\) \(\dfrac{7}{4}\)            = \(\dfrac{4}{5}\)

y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)

y = \(\dfrac{16}{35}\)

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 1:

+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)

\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)

+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)

\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)

\(y=\dfrac{9}{50}\)

Bài 2:

+) \(=\dfrac{2}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{7}{7}=\dfrac{2}{5}\)

+) \(\dfrac{2}{9}:\dfrac{2}{3}:\dfrac{3}{9}\)

\(\dfrac{2}{9}\times\dfrac{3}{2}\times\dfrac{9}{3}=1\)

viết rứa ai mà biết

\(ĐK:x\ne1\\ PT\Leftrightarrow\left\{{}\begin{matrix}x-1=\dfrac{3\cdot4}{3}=4\\y+2=\dfrac{3\cdot8}{4}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\left(tm\right)\\y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)

\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)

\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)

 \(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)

         y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)

         y = \(\dfrac{245}{64}\)

2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)

\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)

\(\dfrac{12}{5}\): y        = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)

 \(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)

        y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)

        y = \(\dfrac{15}{13}\)

\(\dfrac{12}{5}\) - 1\(\dfrac{2}{5}\) \(\times\) y = 1\(\dfrac{1}{4}\)

 \(\dfrac{12}{5}\) - \(\dfrac{7}{5}\) \(\times\) y  = \(\dfrac{5}{4}\)

           \(\dfrac{7}{5}\) \(\times\) y  = \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)

            \(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{23}{20}\)

                   y = \(\dfrac{23}{20}\) : \(\dfrac{7}{5}\)

                   y = \(\dfrac{23}{28}\)

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

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

b:

ĐKXĐ: x<>0

 \(\dfrac{2}{x}+\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)

=>\(6\left(6+xy\right)=3x\)

=>\(x=2\left(6+xy\right)=12+2xy\)

=>\(x\left(1-2y\right)=12\)

mà x,y là các số nguyên

nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)

c: ĐKXĐ: y<>-1

\(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)

=>\(\dfrac{xy+x+3}{3\left(y+1\right)}=\dfrac{1}{6}\)

=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)

=>\(2xy+2x+6=y+1\)

=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)

=>\(\left(2x-1\right)\left(y+1\right)=-6\)

mà x,y là các số nguyên

nên \(\left(2x-1;y+1\right)\in\left\{\left(1;-6\right);\left(-1;6\right);\left(3;-2\right);\left(-3;2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(0;5\right);\left(2;-3\right);\left(-1;1\right)\right\}\)