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

Có \(\frac{a}{b}=\frac{c}{d}\)

=> \(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}\)

Áp dụng dãy tỉ số bằng nhau

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

=> \(\frac{a}{c}=\frac{3a+b}{3c+d}\)

=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

=> Đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b\times k\) ; \(c=d\times k\)

Ta có :

\(\frac{a}{3a+b}=\frac{b\times k}{3\times b\times k+b}=\frac{b\times k}{b\left(3k+1\right)}=\frac{k}{3k+1}\)      (1)

\(\frac{c}{3c+d}=\frac{d\times k}{3\times d\times k+d}=\frac{d\times k}{d\left(3k+1\right)}=\frac{k}{3k+1}\)      (1)

Từ (1) và (2) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

k mk nha bạn !

\(\text{Theo bài ra ta có : }\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{b}=\frac{3a}{3c}\)

\(\text{Áp dụng tính chất của dãy tỷ số banwgf nhau ta có:}\)

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

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

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

\(\Rightarrow\frac{a}{3a+b}=\frac{c}{3a+d}\left(\text{đ}pcm\right)\)

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

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

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

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

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

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

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

<=>\(ad=bc\)

<=>\(\frac{a}{b}=\frac{c}{d}\)(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

Do \(\frac{a}{b}=\frac{c}{d}\) => \(\frac{b}{a}=\frac{d}{c}\)

=> \(3+\frac{b}{a}=3+\frac{d}{c}\)

=> \(\frac{3a+b}{a}=\frac{3c+d}{c}\)

=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\left(đpcm\right)\)

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

ad = bc

3ac + ad = 3ac + bc

a(3c + d) = c(3a + b)

\(\frac{a}{3a+b}=\frac{c}{3c+d}\left(\text{đ}pcm\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)

Xét VT \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)

Xét VP \(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)

Từ (1) và (2) =>Đpcm

Áp dụng tính chất dãy tỉ số bằng nhau, ta có :

     \(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3b+3c+3d+3a}=\frac{a+b+c+d}{3\cdot\left(b+c+d+a\right)}=\frac{1}{3}\)

Do đó :

       \(\frac{a}{3b}=\frac{1}{3}\Rightarrow\frac{a}{b}.\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{a}{b}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow a=b\)

       \(\frac{b}{3c}=\frac{1}{3}\Rightarrow\frac{b}{c}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{b}{c}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow b=c\)

       \(\frac{c}{3d}=\frac{1}{3}\Rightarrow\frac{c}{d}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{c}{d}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow c=d\)

       \(\frac{d}{3a}=\frac{1}{3}\Rightarrow\frac{d}{a}\cdot\frac{1}{3}=\frac{1}{3}\Rightarrow\frac{d}{a}=\frac{1}{3}:\frac{1}{3}=1\Rightarrow d=a\)

\(\Rightarrow a=b=c=d\)

Ta có: \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow a=bk,c=dk\)

\(\frac{3a+7b}{3c-7d}=\frac{3bk+7b}{3dk+7d}=\frac{b\left(3k+7\right)}{d\left(3k+7\right)}=\frac{b}{d}\)(1)

\(\frac{3a-7b}{3c-7d}=\frac{3bk-7b}{3dk-7d}=\frac{b\left(3k-7\right)}{d\left(3k-7\right)}=\frac{b}{d}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{3a+7b}{3c+7d}=\frac{3a-7b}{3c-7d}\)