Question

Tìm x,y biết:

\(\dfrac{x}{2}=\dfrac{y}{5}\) và \(x^2+y^2=70\)

Answer 1

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{25}\)

Áp dụng tc dãy tỉ số bằng nhau :

\(\dfrac{x^2}{4}=\dfrac{y^2}{25}=\dfrac{x^2+y^2}{4+25}=\dfrac{70}{29}\)

Khi đó: \(\left\{{}\begin{matrix}\dfrac{x^2}{4}=\dfrac{70}{29}\\\dfrac{y^2}{25}=\dfrac{70}{29}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=\sqrt{\dfrac{280}{29}}\\x=-\sqrt{\dfrac{280}{29}}\end{matrix}\right.\\\left\{{}\begin{matrix}y=\sqrt{\dfrac{1750}{29}}\\y=-\sqrt{\dfrac{1750}{29}}\end{matrix}\right.\end{matrix}\right.\)

Vậy....

Answer 2

\(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{5^2}\)

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

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

\(\dfrac{x^2}{4}=\dfrac{y^2}{25}=\dfrac{x^2+y^2}{4+25}=\dfrac{70}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{70}{29}=\dfrac{280}{29}\\y^2=25.\dfrac{70}{29}=\dfrac{1750}{29}\end{matrix}\right.\)

Còn lại thì xét mỗi x,y ra 2 giá trị

p/s : Bài này sao số nó "xấu" dữ z ???

Answer 3

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\)

=>\(\left\{{}\begin{matrix}x=2.k\\y=5.k\end{matrix}\right.\)

Ta có: x² + y²=70

=>(2.k)²+(5.k)²=70

=>4.k²+25.k²=70

=>k².(25+4)=70

=>k².29=70

=>k²=\(\dfrac{70}{29}\)

=>k=\(\pm\dfrac{70}{29}\)

Với k=\(\dfrac{70}{29}\) => \(\left\{{}\begin{matrix}x=2.\dfrac{70}{29}=\dfrac{140}{29}\\y=5.\dfrac{70}{29}=\dfrac{350}{29}\end{matrix}\right.\)

Với k=\(-\dfrac{70}{29}\) => \(\left\{{}\begin{matrix}x=2.\left(-\dfrac{70}{29}\right)=-\dfrac{140}{29}\\y=5.\left(-\dfrac{70}{29}\right)=-\dfrac{350}{29}\end{matrix}\right.\)

Vậy +với k=\(\dfrac{70}{29}\) => \(\left\{{}\begin{matrix}x=2.\dfrac{70}{29}=\dfrac{140}{29}\\y=5.\dfrac{70}{29}=\dfrac{350}{29}\end{matrix}\right.\)

+Với k=\(-\dfrac{70}{29}\) => \(\left\{{}\begin{matrix}x=2.\left(-\dfrac{70}{29}\right)=-\dfrac{140}{29}\\y=5.\left(-\dfrac{70}{29}\right)=-\dfrac{350}{29}\end{matrix}\right.\)

Answer 4

\(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow2x=5y\Leftrightarrow x=\dfrac{2}{5}y\)

\(x=\dfrac{2}{5}y\Leftrightarrow x^2=\left(\dfrac{2}{5}\right)^2y^2\Leftrightarrow x^2=\dfrac{4}{25}y^2\)

\(x^2=70:\left(4+25\right).4=\dfrac{280}{29}\)

\(\sqrt{\dfrac{280}{29}};-\sqrt{\dfrac{280}{29}}\)

\(y^2=70:\left(4+25\right).25=\dfrac{1750}{29}\)

\(\Leftrightarrow y=\sqrt{\dfrac{1750}{29}};-\sqrt{\dfrac{1750}{29}}\)