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Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\)

Ta có:

\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2\left(bk\right)^2-2bkb+5b^2}{2b^2+3bkb}=\dfrac{2b^2k^2-2b^2k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(1\right)\)

\(\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(2\right)\)

Từ (1) và (2) suy ra:

\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)

Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2b^2k^2-3b^2k+5d^2}{2b^2+3b^2k}\)

\(=\dfrac{b^2k\left(2k-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{k\left(2k-3+5\right)}{2+3k}\) (1)

\(\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}\)

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

Từ (1), (2) \(\Rightarrow\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\left(đpcm\right)\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow a=bk;c=dk\)

\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+5}{3k+2}\)

\(\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\frac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\frac{2k^2-3k+5}{3k+2}\)

nên 2 phân số trên bằng nhau (đpcm)

Đặt: \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có : \(\frac{2a^2-3ab+5b^2}{2b^2+3ab}\)

<=> \(\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}\)

<=> \(\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}\)

<=> \(\frac{2k^2-3k+5}{2+3k}\left(1\right)\)

Ta có: \(\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)

<=> \(\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}\)

<=> \(\frac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}\)

<=> \(\frac{2k^2-3k+5}{2+3k}\left(2\right)\)

Từ 1 và 2 => đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)

\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2\left(bk^2\right)-3bkb+5b^2}{2b^2+3bkb}=\frac{2b^2.k^2-3kb^2+5b^2}{2b^2+3b^2.k}\)\(=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+5}{2+3k}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)\(=\frac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3dkd}\)

Tương tự nhóm tiếp là ra

=>bằng nhau

Điều kiện mẫu thức xác định là sao?

đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Thay a và c vào VP và VT sẽ bằng nhau

Ai help me vs