Question

Tìm x, y biết : \(\dfrac{3}{5}x=\dfrac{2}{3}y\) và \(x^2-y^2=38\)

Answer 1

Ta có:

3/5x=2/3y(1) , x^2-y^2=38(2)

(1)=> (9/10) x . thay vào (1) =>x^2 -((9/10x))^2=38<=>x^2-(81/100)x^2=38

<=> (19/100)x^2=38<=>x^2=(38/19)*100=200

<=> x=10 can 2, y=(9/10)x=9can2

<=> X=-10can2, y=(9/10)x=-9can2

Answer 2

\(\dfrac{3}{5}x=\dfrac{2}{3}y\Rightarrow\dfrac{x}{\dfrac{5}{3}}=\dfrac{y}{\dfrac{3}{2}}\Rightarrow\left(\dfrac{x}{\dfrac{5}{3}}\right)^2=\left(\dfrac{y}{\dfrac{3}{2}}\right)^2\)

\(\Rightarrow\dfrac{x^2}{\dfrac{25}{9}}=\dfrac{y^2}{\dfrac{9}{4}}\)

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

\(\dfrac{x^2}{\dfrac{25}{9}}=\dfrac{y^2}{\dfrac{9}{4}}=\dfrac{x^2-y^2}{\dfrac{25}{9}-\dfrac{9}{4}}=\dfrac{28}{\dfrac{19}{36}}\)

Áp dụng tính