Question

Cho a/b = c/d với a, b, c, d > 0. Chứng minh rằng \(\dfrac{3a^2+10b^20-ab}{7a^2+b^2+5ab}=\dfrac{3c^2+10d^2-cd}{7c^2+d^2+5cd}\)

Answer 1

không cs số 0 đâu 

Answer 2

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

Sửa: \(\dfrac{3a^2+10b^2-ab}{7a^2+b^2+5ab}=\dfrac{3b^2k^2+10b^2-b^2k}{7b^2k^2+b^2+5b^2k}=\dfrac{b^2\left(3k^2+10-k\right)}{b^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(1\right)\)

\(\dfrac{3c^2+10d^2-cd}{7c^2+d^2+5cd}=\dfrac{3d^2k^2+10d^2-d^2k}{7d^2k^2+d^2+5d^2k}=\dfrac{d^2\left(3k^2+10-k\right)}{d^2\left(7k^2+1+5k\right)}=\dfrac{3k^2+10-k}{7k^2+1+5k}\left(2\right)\)

\(\left(1\right)\left(2\right)\RightarrowĐpcm\)