Question

Bài 4: Chứng minh rằng:

Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì:

a) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

b)\(\dfrac{7a^2+3ab}{11b^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

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Answer 1

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

a) Từ (*) ta có:

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

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

Từ (1) và (2) suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

b) Từ (*) ta có:

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\) (3)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\) (4)

Từ (3) và (4) suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

P/s: test lại đề phần b), mẫu số của vế trái

Answer 2

a, Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Vì \(\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{5a}{5c}=\dfrac{3b}{3d}\)\(=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(\Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\left(đpcm\right)\)

Vậy \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+5d}{5c-5d}\)