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

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

\(\dfrac{2}{5}\) 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 = \(\dfrac{23}{28}\)

Phương trình đâu bạn ?

$x=144,$ $y=36$.

a.

1x;b=1y.

⎧⎪ ⎪⎨⎪ ⎪⎩a+b=56
2416.a+15.b=215⇔{a=12b=14

⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1x=121y=13

⇔{x=2y=3
c,

xy−xy+12=1xy−12−xy=2

xy=a;xy+12=b;xy−12=c

1b+1c=2a(1)

1b−1a)+(1c−1a)=0

a−bab+a−cac=0⇔1ab−2ac=0

1a(1b−2c)=0⇒1b=2c⇒c=2b

$\Rightarrow \{\begin{array}{c}a-b=1\\ 2b-a=2\end{array}\iff \{\begin{array}{c}a=4\\ b=3\end{array}$

⇒⎧⎪ ⎪⎨⎪ ⎪⎩xy=4xy+12=3⇔⎧⎪⎨⎪⎩x=4y4yy+12=3⇔{x=144y=36

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) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

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

=> 2(2x+1) = 6.7

4x+2=42

4x=40

x=10

Vậy x=10

a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\\ =>6.7=2.\left(2x+1\right)\\ =>2x+1=\dfrac{6.7}{2}=\dfrac{42}{2}=21\\ =>2x=21-1=20\\ =>x=\dfrac{20}{2}=10\)

b) \(\dfrac{24}{7x-3}=-\dfrac{4}{25}\\ =>24.25=-4.\left(7x-3\right)\\ =>7x-3=\dfrac{24.25}{-4}=-150\\ =>7x=-150+3=-147\\ =>x=\dfrac{-147}{7}=-21\)

c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=-\dfrac{12}{18}\\ =>x-6=\dfrac{4.18}{-12}=-6\\ =>x=-6+6=0\\ y=\dfrac{-12.24}{18}=-16\)

d) \(-\dfrac{1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\\ < =>-\dfrac{8}{40}\le-\dfrac{5x}{40}\le\dfrac{10}{40}\\ =>-8\le-5x\le10\\ Mà:-8< -5.1< -5.0< -5.\left(-1\right)< -5.\left(-2\right)=10\\ =>x\in\left\{-2;-1;0;1\right\}\)

e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\\ < =>\dfrac{x+46}{20}=\dfrac{5x+2}{5}\\ =>5\left(x+46\right)=20\left(5x+2\right)\\ < =>5x+230=100x+40\\ < =>230-40=100x-5x\\ < =>190=95x\\ =>x=\dfrac{190}{95}=2\)

f) \(y\dfrac{5}{y}=\dfrac{56}{y}\\ < =>\dfrac{y^2+5}{y}=\dfrac{56}{y}\\ =>y\left(y^2+5\right)=56y\\ =>y^2+5=\dfrac{56y}{y}=56\\ =>y^2=56-5=51\\ =>y=\sqrt{51}\)

\(\dfrac{24}{7x-3}=\dfrac{-4}{25}\)

\(\Rightarrow\left(7x-3\right).\left(-4\right)=24.25\)

\(\Rightarrow-28x+12=600\)

\(\Rightarrow-28x=600-12=588\)

\(\Rightarrow\)\(x=588:\left(-28\right)=-21\)

Vậy \(x=-21\)

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

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

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   = \(\dfrac{23}{48}\)