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

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

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

a)=2

b)=1/3

a,2

b,1/3

a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)

\(\dfrac{3}{10}-x=\dfrac{7}{10}\)

x = \(\dfrac{3}{10}-\dfrac{7}{10}\)

x=\(\dfrac{-4}{10}\)

b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)

\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)

\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)

\(x=\dfrac{-64}{9}\)

c)=>2.18=(x-3).(x-3)

=>36=(x-3)\(^2\)

=>6\(^2\)=(x-3)\(^2\)

6= x-3

x=6+3=9

ĐKXĐ: \(x\ne1;y\ne1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-1}=u\\\dfrac{1}{y-1}=v\end{matrix}\right.\) hệ trở thành:

\(\left\{{}\begin{matrix}5u+v=10\\u-3v=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}15u+3v=30\\u-3v=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}16u=48\\v=\dfrac{u-18}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=3\\v=-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=3\\\dfrac{1}{y-1}=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=\dfrac{1}{3}\\y-1=-\dfrac{1}{5}\end{matrix}\right.\)

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

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

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

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

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

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

\(\dfrac{y}{3}=2\Rightarrow y=6\)

\(\dfrac{z}{7}=2\Rightarrow z=14\)

\(\left(0.25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)
\(=0.4^{10}+\sqrt{25-9}\)
\(=0+\sqrt{16}=0+4=4\)

\(\dfrac{5}{20}+\dfrac{18}{11}-25\%-\left(\dfrac{18}{11}-\dfrac{4}{9}\right)\)
\(=\dfrac{5}{20}+\dfrac{18}{11}-\dfrac{1}{4}-\dfrac{18}{11}+\dfrac{4}{9}\)
\(=\left(\dfrac{5}{20}-\dfrac{1}{4}\right)+\left(\dfrac{18}{11}-\dfrac{18}{11}\right)+\dfrac{4}{9}\)
\(=0+0+\dfrac{4}{9}=\dfrac{4}{9}\)

    (0,25)¹⁰.4¹⁰ + \(\sqrt{5^2-3^2}\)

=   (0,25 .4)¹⁰ + \(\sqrt{25-9}\)

= 1¹⁰ + \(\sqrt{16}\)

= 1 + 4 

= 5

\(\dfrac{5}{20}\) + \(\dfrac{18}{11}\) - 25% - ( \(\dfrac{18}{11}\) - \(\dfrac{4}{9}\))

= 25% +  \(\dfrac{18}{11}\) - 25% - \(\dfrac{18}{11}\) + \(\dfrac{4}{9}\)

= ( 25% - 25%)  + ( \(\dfrac{18}{11}\) - \(\dfrac{18}{11}\)) + \(\dfrac{4}{9}\)

= 0  + 0 + \(\dfrac{4}{9}\)

= \(\dfrac{4}{9}\)

Đặt \(\dfrac{1}{y-1}=a\), hpt tở thành

\(\left\{{}\begin{matrix}\dfrac{5}{x+1}+a=10\\\dfrac{1}{x-2}+3a=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{x+1}+3a=30\left(1\right)\\\dfrac{1}{x-1}+3a=18\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)-\left(2\right)\), ta được:

\(\dfrac{15}{x+1}-\dfrac{1}{x-1}=12\\ \Leftrightarrow\dfrac{15x-15-x-1}{\left(x-1\right)\left(x+1\right)}=12\\ \Leftrightarrow12x^2-12=14x-16\\ \Leftrightarrow12x^2-14x+4=0\\ \Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Với \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{10}{3}+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{10y-7}{3\left(y-1\right)}=10\)

\(\Leftrightarrow30y-30=10y-7\Leftrightarrow y=\dfrac{23}{20}\)

Với \(x=\dfrac{2}{3}\Leftrightarrow3+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{1}{y-1}=7\Leftrightarrow7y-7=1\Leftrightarrow y=\dfrac{8}{7}\)

Vậy \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)