\(\dfrac{4}{x-3}=\dfrac{8}{y-6}=\dfrac{20}{z-15}\)

\(\Rightarrow\dfrac{x-3}{4}=\dfrac{y-6}{8}=\dfrac{z-15}{20}\)

\(\Rightarrow\dfrac{x}{4}-\dfrac{3}{4}=\dfrac{y}{8}-\dfrac{3}{4}=\dfrac{z}{20}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{8}=\dfrac{z}{20}\)

Đặt: \(\dfrac{x}{4}=\dfrac{y}{8}=\dfrac{z}{20}=k\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=8k\\z=20k\end{matrix}\right.\)

Thay vào đk đề bài: \(640k^3=640\Leftrightarrow k=1\)

Vậy \(\left\{{}\begin{matrix}x=4\\y=8\\z=20\end{matrix}\right.\)

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

Ko đề cho thêm \(\dfrac{20}{4²}\) mà 

a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)

\(B=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{25}{75}\cdot\dfrac{26}{8}=\dfrac{13}{12}\)

b: Để A<x<B thì -11/12<x<13/12

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)

=>-x-6-3=4

=>-x-9=4

=>-x=5

hay x=-5

b: =>(x+1)²=16

=>x+1=4 hoặc x+1=-4

=>x=3 hoặc x=-5

c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)

=>x-29=0

hay x=29

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

\(\Leftrightarrow\left(x+2\right).2=\left(2x+1\right).0,5\)

\(\Leftrightarrow2x+4=x+0,5\)

\(\Leftrightarrow x=-3,5\)

Vậy...

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

\(\Leftrightarrow\dfrac{4.\left(x+2\right)}{2}=\dfrac{2x+1}{2}\)

\(\Rightarrow4x+8=2x+1\)

\(\Leftrightarrow4x-2x=1-8\)

\(\Leftrightarrow2x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(x=\dfrac{-7}{2}\)

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

\(\Rightarrow\dfrac{4\left(x+2\right)}{2}=\dfrac{2x+1}{2}\)

\(\Rightarrow\dfrac{4x+8}{2}=\dfrac{2x+1}{2}\)

\(\Rightarrow4x+8=2x+1\)

\(\Rightarrow4x+2x=8+1\)

\(\Rightarrow6x=9\)

\(\Rightarrow x=\dfrac{9}{6}=\dfrac{3}{2}\)

Vậy \(x=\dfrac{3}{2}\)

\(\dfrac{6}{13}.\dfrac{8}{7}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{6}{13}.\left(\dfrac{8}{7}+\dfrac{9}{7}\right)-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{6}{13}.\dfrac{17}{7}-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{102}{91}-\dfrac{18}{91}=\dfrac{12}{13}\)

a, \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{13}{12}+\dfrac{13}{32}=\dfrac{143}{96}\)

b, \(\dfrac{5}{-8}+\dfrac{14}{39}-\dfrac{6}{10}\)

\(\dfrac{-5}{8}+\dfrac{14}{39}-\dfrac{3}{5}\)

\(=\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{14}{39}\)

\(=\dfrac{-49}{40}+\dfrac{14}{39}=\dfrac{-1351}{1560}\)

= 28/15 . 3/4 - ( 11/20 + 1/4 ) : 7/3

= 28/15 . 3/4 - 4/5 : 7/3

= 7/5 - 12/35

= 37/35

= \(\dfrac{28}{15}\) . \(\dfrac{3}{4}\) - (\(\dfrac{11}{20}\) + \(\dfrac{1}{4}\)) : \(\dfrac{7}{3}\)

= \(\dfrac{7}{5}\) - (\(\dfrac{11}{20}\) + \(\dfrac{5}{20}\)) : \(\dfrac{7}{3}\)

=  \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) : \(\dfrac{7}{3}\)

=  \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) x \(\dfrac{3}{7}\)

=  \(\dfrac{7}{5}\) - \(\dfrac{12}{35}\)

=  \(\dfrac{49}{35}\) - \(\dfrac{12}{35}\)

= \(\dfrac{37}{35}\)

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)