C₁: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a=bk;c=dk\)

\(\Rightarrow\dfrac{3a-4b}{b}=\dfrac{bk-4b}{b}=\dfrac{b\left(k-4\right)}{b}=k-4\left(1\right)\)

\(\Rightarrow\dfrac{3c-4d}{d}=\dfrac{dk-4d}{d}=\dfrac{d\left(k-4\right)}{d}=k-4\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{3a-4b}{b}=\dfrac{3c-4d}{d}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{3a-4b}{b}=\dfrac{3\cdot bk-4b}{b}=3k-4\)

\(\dfrac{3c-4d}{d}=\dfrac{3dk-4d}{d}=3k-4\)

Do đó: \(\dfrac{3a-4b}{b}=\dfrac{3c-4d}{d}\)

bạn tham khảo nhé!

a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a+4b}{3c+4d}=\frac{3a-4b}{3c-4d}.\)

\(\Rightarrow\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\)

b) ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{2c}{2d}=\frac{4a}{4b}\)

Lại có: \(\frac{5a}{5b}=\frac{2c}{2d}=\frac{5a+2c}{5b+2d}\)

\(\Rightarrow\frac{4a}{4b}=\frac{5a+2c}{5b+2d}\Rightarrow\frac{5a+2c}{4a}=\frac{5b+2d}{4b}\)

c) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Lại có: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)

\(\Rightarrow\frac{\left(a+b^2\right)}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)

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