a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)

2: y \(\times\) \(\dfrac{3}{5}\) = \(\dfrac{9}{10}\)

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

2: y = \(\dfrac{3}{2}\)

    y = 2 : \(\dfrac{3}{2}\)

     y = \(\dfrac{4}{3}\)

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

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

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

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

            y = \(\dfrac{8}{5}\)

\(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{7}{2}\) - y) = \(\dfrac{3}{2}\)

           \(\dfrac{7}{2}\) - y = \(\dfrac{3}{2}\) : \(\dfrac{3}{4}\)

            \(\dfrac{7}{2}\) - y = 2

                   y = \(\dfrac{7}{2}\) - 2

                   y = \(\dfrac{3}{2}\)

1/2-2y=9/20

=>2y=1/2-9/20=1/20

=>y=1/20:2=1/40

b,3/5:4/3:y=2+7/10=9/20:y=27/10

=>y=9/20:27/10=1/6

c,y+y*3/2-y*1/2=1/10

=>y(1+3/2-1/2)=1/10

=>2y=1/10

=>y=1/10:2=1/20

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

\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)

\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)

a) \(...\Rightarrow\dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\)

\(\Rightarrow y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\Rightarrow y:\dfrac{11}{4}=\dfrac{24}{10}-\dfrac{15}{10}\)

\(\Rightarrow y:\dfrac{11}{4}=\dfrac{9}{10}\Rightarrow y=\dfrac{9}{10}x\dfrac{11}{4}=\dfrac{99}{40}\)

b) \(...\Rightarrow\dfrac{5}{4}+\dfrac{11}{5}xy=\dfrac{13}{5}\Rightarrow\dfrac{11}{5}xy=\dfrac{13}{5}-\dfrac{5}{4}\)

\(\Rightarrow\dfrac{11}{5}xy=\dfrac{52}{20}-\dfrac{25}{20}\Rightarrow\dfrac{11}{5}xy=\dfrac{27}{20}\)

\(\Rightarrow y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{20}x\dfrac{5}{11}=\dfrac{27}{44}\)

c) \(...\Rightarrow\dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\Rightarrow\dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{9}{4}:y=\dfrac{56}{20}-\dfrac{15}{20}\Rightarrow\dfrac{9}{4}:y=\dfrac{39}{20}\)

\(\Rightarrow y=\dfrac{9}{4}:\dfrac{39}{20}\Rightarrow y=\dfrac{9}{4}x\dfrac{20}{39}=\dfrac{15}{13}\)

d) \(...\Rightarrow x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\Rightarrow x:\dfrac{10}{3}=\dfrac{24}{10}+\dfrac{7}{10}\)

\(\Rightarrow x:\dfrac{10}{3}=\dfrac{31}{10}\Rightarrow x=\dfrac{31}{10}x\dfrac{10}{3}=\dfrac{31}{3}\)

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

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

a: \(y+\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{3}{4}\)

=>\(y+\dfrac{1}{6}=\dfrac{3}{4}\)

=>\(y=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)

b: \(y+\dfrac{1}{2}\cdot\dfrac{1}{5}=\dfrac{3}{10}\)

=>\(y+\dfrac{1}{10}=\dfrac{3}{10}\)

=>\(y=\dfrac{3}{10}-\dfrac{1}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)

tìm X ạ

a) \(\dfrac{x}{x-y}+\dfrac{2y^2}{x^2-y^2}-\dfrac{x}{x+y}=\dfrac{x\left(x+y\right)+2y^2-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x^2+xy+2y^2-x^2+xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y^2+2xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y}{x-y}\)

b) \(B=\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}=\dfrac{x\left(x+2\right)-4x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)

c) \(\dfrac{5}{x+1}-\dfrac{10}{-x^2+x-1}-\dfrac{15}{x^3+1}=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3+1}=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

Do đó: x=16; y=24; z=30