Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow a=bk;c=dk\)
\(VP=\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2\left(bk\right)^2-3bkb+5b^2}{2b^2+3bkb}=\dfrac{2b^2.k^2-2b^2.k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(1\right)\)
\(VT=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2\left(dk\right)^2-3dkd+5d^2}{2\left(dk\right)^2+3dkd}=\dfrac{2.d^2.k^2-3d^2.k+5.d^2}{2.d^2.k^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\)\(\Rightarrow\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\)\(\Rightarrow\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)
\(\Rightarrowđpcm\)
