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Đặt a/b=c/d=k rồi thay vào nha bạn

GỢI Ý:

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

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

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

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

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

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

Điều kiện mẫu thức xác định là sao?

Ta có : \(\frac{a}{b}=\frac{c}{d}\)=> \(\frac{a}{c}=\frac{b}{d}\)

Đặt \(\frac{a}{c}=\frac{b}{d}=k\)=> \(\hept{\begin{cases}a=ck\\d=dk\end{cases}}\)

Khi đó, ta có : \(\frac{2\left(ck\right)^2-3\left(ck\right)\left(dk\right)+5\left(dk\right)^2}{2\left(dk\right)^2+3\left(ck\right)\left(dk\right)}=\frac{2c^2k^2-3cdk^2+5d^2k^2}{2d^2k^2+3cdk^2}=\frac{\left(2c^2-3cd+5d^2\right)k^2}{\left(2d^2+3cd\right)k^2}\)

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