Mình hướng dẫn thôi. Chứ giờ đang bận.
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\).Rồi thay a = kb; c=kd vào từng vế. Thấy hai vế bằng nhau => đpcm
\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}=\frac{2a^2-3ab+5b^2}{2b^2-3cd+5d^2}=\frac{2b^2+3ab}{2d^2+3cd}\)
\(=>\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)
Đặt\(\frac{a}{b}=\frac{c}{d}=k\),ta có:
\(a=bk\)\(c=dk\)\(\Rightarrow\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2\left(bk\right)^2-3bkb+5b^2}{2b^2+3bkb}=\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\frac{b^2.\left(2k^2+3k+5\right)}{b^2.\left(2+3k\right)}\)\(=\frac{2k^2+3k+5}{2+3k}\left(1\right)\)
\(\Rightarrow\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\frac{d^2.\left(2k^2+3k+5\right)}{d^2.\left(2+3k\right)}\)
\(=\frac{2k^2+3k+5}{2+3k}\)(2)
Từ (1) và (2) suy ra:
\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)(đpcm)
