Question

Tìm x,y biết

4x=3y và (x-y)²+(x+y)²=50

 

 

Answer 1

Theo đầu bài: \(\left(x-y\right)^2+\left(x+y\right)^2=50\)

=>\(x^2-2xy+y^2+x^2+2xy+y^2=50\)

=>\(2x^2+2y^2=50\)

=>\(2\left(x^2+y^2\right)=50\)

=>\(x^2+y^2=50:2\)

=>\(x^2+y^2=25\) (*)

Xét \(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\)

Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)

Dựa vào (*) ta có: x²+y²=25

=>\(\left(3k\right)^2+\left(4k\right)^2=25\)

=>\(9k^2+16k^2=25\)

=>\(k^2\left(9+16\right)=25\)

=>\(k^2.25=25\)

=>\(k^2=25:25=1\)

=>x=-1 hoặc x=1

+)Nếu x=-1

=>\(\hept{\begin{cases}x=\left(-1\right).3=-3\\y=\left(-1\right).4=-4\end{cases}}\)

+)Nếu x=1

=>\(\hept{\begin{cases}x=1.3=3\\y=1.4=4\end{cases}}\)

Vậy có 2 cặp x;y thỏa mãn là ...

Answer 2

\(4x=3y\Leftrightarrow x=\frac{3y}{4}\)(*)

Thay (*) vào \(\left(x-y\right)^2+\left(x+y\right)^2=50\Leftrightarrow\left(\frac{3y}{4}-y\right)^2+\left(\frac{3y}{4}+y\right)=50\Leftrightarrow\left(\frac{3y-4y}{4}\right)^2+\left(\frac{3y+4y}{4}\right)^2=50\)

\(\Leftrightarrow\frac{y^2}{16}+\frac{49y^2}{16}=50\Leftrightarrow49y^2+y^2=800\Leftrightarrow50y^2=800\Leftrightarrow y^2=16\Leftrightarrow y=4\)