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Ta có:
a/b=c/d => a/c=b/d=2a/2c=3b/3d
= 2a+3b/2c+3d=2a-3b/2c-3d
=> 2a+3b/2a-3b=2c+3d/2c-3d (ĐPCM)
ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{d}{b}=\frac{c}{a}\Rightarrow\frac{c+d}{a+b}\Rightarrow\frac{3c+3d}{3a+3b}=\frac{3c-3d}{3a-3b}\)
\(\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)\(\left(điềuphảichứngminh\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{3a+5b}{3a-5b}=\frac{3bk+5b}{3bk-5b}=\frac{b\left(3k+5\right)}{b\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
\(\frac{3c+5d}{3c-5d}=\frac{3dk+5d}{3dk-5d}=\frac{d\left(3k+5\right)}{d\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
Vậy từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
\(a,\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{c}+1=\frac{b}{d}+1\)
\(\Rightarrow\frac{a}{c}+\frac{c}{c}=\frac{b}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+c}{c}=\frac{b+d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{5b^2}{5d^2}=\frac{2a^2}{2c^2}=\frac{7ab}{7cd}\)
\(=\frac{3a^2+5b^2}{3c^2+5d^2}=\frac{2a^2+7ab}{2c^2+7cd}\) ( tích chất dãy tỉ số bằng nhau )
\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
=> (3a + 5b)(3c - 5d) = (3a - 5b)(3c + 5d)
=> 9ac - 15ad + 15bc - 25bd = 9ac + 15ad - 15bc - 25bd
=> 9ac - 15ad + 15bc - 25bd - (9ac + 15ad - 15bc - 25bd) = 0
=> 9ac - 15ad + 15bc - 25bd - 9ac - 15ad + 15bc + 25bd = 0
=> (9ac - 9ac) + (-15ad - 15ad) + (15bc + 15bc) + (-25bd + 25bd) = 0
=> -30ad + 30bc = 0
=> -30ad = -30bc
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\)