Đề sai :(
Sửa:
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\left(a;b;c;d\ne0\right).\)
Chứng tỏ rằng: \(\dfrac{2a-5b}{3a}=\dfrac{2c-5d}{3c}.\)
Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk.\)
Ta có:
\(\dfrac{2a-5b}{3a}=\dfrac{2bk-5b}{3bk}=\dfrac{2.5.b\left(k-1\right)}{3bk}=\dfrac{10\left(k-1\right)}{3k}_{\left(1\right)}.\)
\(\dfrac{2c-5d}{3c}=\dfrac{2dk-5d}{3dk}=\dfrac{2.5.d\left(k-1\right)}{3dk}=\dfrac{10\left(k-1\right)}{3k}_{\left(2\right)}.\)
Từ \(_{\left(1\right)\&\left(2\right)}\Rightarrow\dfrac{2a-5b}{3a}=\dfrac{2c-5d}{3c}.\)
\(\Rightarrowđpcm.\)
chỗ kia mình viết sai nha đề đúng là\(\dfrac{2a-5b}{3a}\)=\(\dfrac{2c-5d}{3c}\)
