Question

[Huyền Thụn]

1, Cho \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\). CMR \(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}\) = \(\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)

Answer 1

[Nguyễn Thanh Hằng]

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 \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Answer 2

[Huyền Thụn]

thôi mk làm đc rùi ko cần nữa nha ^-^