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

mik làm nhiều rùi quá dễ

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)

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

=> (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

=> \(\frac{a}{b}=\frac{c}{d}\)

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

mà \(\frac{a}{c}=\frac{3a}{3c}\)

\(\Rightarrow\frac{2a-5b}{2c-5d}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

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

Mà \(\frac{a}{c}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{2c-5d}=\frac{3a}{3c}\Rightarrow\frac{2a-5b}{3a}=\frac{2c-5d}{3c}\)

(ĐPCM)

MK LÀ NGƯỜI TRẢ LỜI ĐẦU TIÊN NHA MẤT MÔT HỒI MỚI NGHĨ RA

\(a,\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{a}{c}+1=\frac{b}{d}+1\)

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

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

Ta có :

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

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

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

\(\frac{3a+5b}{2a-b}=\frac{3c+5d}{2c-d}\)

<=>\(\left(3a+5b\right)\left(2a-b\right)=\left(3c+5d\right)\left(2c-d\right)\)

<=>\(6ac+10ad-3bc-5bd=6ac+10bc-3ad-5bd\)

<=>\(10ad-3bc=10bc-3ad\)

<=>\(10ad-3bc-10bc+3ad=0\)

<=>\(13ad-13ac=0\)

<=>\(13ad=13ac\)

<=>\(ad=bc\)

<=>\(\frac{a}{b}=\frac{c}{d}\)(đpcm)

Ta có: \(\frac{3a+5b}{2a-b}=\frac{3c+5d}{2c-d}\)

=> (3a+5b)(2c-d) =(2a-b)(3c+5d)

=> 3a(2c-d) +5b(2c-d) =2a(3c+5d) -b(3c+5d)

=> 6ac -3ad +10bc -5bd =6ac +10ad -3bc -5bd

=>7bc=7ad

=> bc=ad 

=> a/b =c/d

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

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