Question

Cho \(\dfrac{x}{y}=\dfrac{z}{t}.\) Chứng minh rằng \(\dfrac{2x^2-3xy+5y^2}{2y^2+3xy}=\dfrac{2z^2-3zt+5t^2}{2t^2+3zt}\) ( x ; y ;z là các BT xác định)

Answer 1

Đặt \(\dfrac{x}{y}=\dfrac{z}{t}=k\Rightarrow x=ky;z=kt\)

Xét \(VT=\dfrac{2x^2-3xy+5y^2}{2y^2+3xy}=\dfrac{2\left(ky\right)^2-3ky\cdot y+5y^2}{2y^2+3ky\cdot y}\)

\(=\dfrac{2k^2y^2-3ky^2+5y^2}{2y^2+3ky^2}=\dfrac{y^2\left(2k^2-3k+5\right)}{y^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{3k+5}\)

Và \(VP=\dfrac{2z^2-3zt+5t^2}{2t^2+3zt}=\dfrac{2\left(kt\right)^2-3kt\cdot t+5t^2}{2t^2+3kt\cdot t}\)

\(=\dfrac{2k^2t^2-3kt^2+5t^2}{2t^2+3kt^2}=\dfrac{t^2\left(2k^2-3k+5\right)}{t^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{3k+5}\)

Dễ thấy \(VT=VP\)\(\forall \frac{x}{y}=\frac{z}{t}\) nên ta có ĐPCM