Question

tìm x

\(\frac{x}{x+4}\)= \(\frac{5}{6}\)

\(\frac{x-3}{x+5}\)=\(\frac{5}{7}\)

\(\frac{x+4}{20}\)=\(\frac{5}{20}\)

\(\frac{x+4}{20}\)=\(\frac{5}{x+4}\)

giúp mình vs

Answer 1

\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)

\(=>6x-5x=20=>x=20\)

Answer 2

áp dụng \(\frac{a}{b}=\frac{c}{d}< =>a.d=b.c\)
 

Answer 3

\(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Rightarrow\left(x+4\right)^2=20.5=100=10^2\)

\(\Rightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)