Question

Tìm x, y biết:

A)\(\dfrac{x+1}{x-2}=\dfrac{x+2}{x-9}\)

B)\(\left[{}\begin{matrix}x\left(x+y\right)=10\\y\left(x+x\right)=6\end{matrix}\right.\)

Answer 1

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+1}{x-2}=\dfrac{x+2}{x-9}=\dfrac{x+1-x-2}{x-2-x+9}=-\dfrac{1}{7}\)

Hay \(\dfrac{x+1}{x-2}=-\dfrac{1}{7}\Leftrightarrow-x+2=7x+7\Leftrightarrow-x=7x+5\Leftrightarrow-x-7x=5\Leftrightarrow-8x=5\Leftrightarrow x=-\dfrac{5}{8}\)b) phải sử dụng \(\left\{{}\begin{matrix}x\left(x+y\right)=10\\y\left(x+y\right)=6\end{matrix}\right.\)(sửa đề)

\(\Leftrightarrow\left(x+y\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+y=4\\x+y=-4\end{matrix}\right.\)

Nên \(\left[{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\)