Question

tìm x trong các tỉ lệ thức sau

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

b) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

Answer 1

a) Ta có:\(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-5x=21+25\)

\(\Rightarrow2x=46\)

\(\Rightarrow x=46:2=23\)

Answer 2

\(a,\dfrac{x-3}{x+5}=\dfrac{5}{7}\\ \Leftrightarrow7x-21=5x+25\\ \Leftrightarrow2x=46\\ \Leftrightarrow x=23\)

Vậy......

\(b,\dfrac{x+4}{20}=\dfrac{5}{x+4}\\ \Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow x+4=\pm10\\ \Leftrightarrow x\in\left\{-14;6\right\}\)

Vậy.........

Answer 3

b) Ta có: \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

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

\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10-4\\x=-10-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)