Question

Tìm số chưa biết

a)\(\dfrac{x}{y}=\dfrac{9}{7};\dfrac{y}{z}=\dfrac{7}{3};x=y-z=-15\)

b)\(\dfrac{x}{-3}=\dfrac{y}{-8};x^2-y^2=-\dfrac{44}{5}\)

Answer 1

a) \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{z}{3}=\dfrac{y}{7}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

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

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y-z}{9-7-3}=\dfrac{-15}{-1}=15\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.9\\y=15.7\\z=15.3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=135\\y=105\\z=45\end{matrix}\right.\)

Vậy, x = 135, y = 105, z = 45

Answer 2

b, \(\dfrac{x}{-3}=\dfrac{y}{-8}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{64}\)

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

\(\dfrac{x^2}{9}=\dfrac{y^2}{64}=\dfrac{x^2-y^2}{9-64}=-\dfrac{44}{\dfrac{5}{-55}}=-\dfrac{44}{5}:\left(-55\right)=-\dfrac{44}{5}.-\dfrac{1}{55}=\dfrac{44}{275}=0,16\)

+) \(\dfrac{x^2}{9}=0,16\Rightarrow x^2=1,44\Rightarrow x=\pm1,2\)

+) \(\dfrac{y^2}{64}=0,16\Rightarrow y^2=10,24\Rightarrow y=\pm3,2\)

Vậy ...