Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)

Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)

c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)

\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)

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

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

=> 9ac + 3ad - 3bc - bd = 9ac - 3ad + 3bc - bd

=> 3ad - 3bc = -3ad + 3bc

=> 3ad + 3ad = 3bc + 3bc

=> 6ad = 6bc

=> ad = bc

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

b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó \(\frac{b^2+d^2}{a^2+c^2}=\frac{b^2+d^2}{\left(bk\right)^2+\left(dk\right)^2}=\frac{b^2+d^2}{d^2k^2+d^2k^2}=\frac{b^2+d^2}{k^2\left(b^2+d^2\right)}=\frac{1}{k^2}\)(1);

\(\frac{bd}{ac}=\frac{bd}{bkdk}=\frac{1}{k^2}\left(2\right)\)

Từ (1)(2) => \(\frac{b^2+d^2}{a^2+c^2}=\frac{bd}{ac}\)(đpcm)

Xét \(\frac{a}{b}=k;\frac{c}{d}=k\)

=> a= bk; c= dk

Thay:

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

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

Ta thấy (1)= (2)

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

theo bài ra ta có:

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

áp dụng t/c dãy tỉ số bằng nhau ta có:

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

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

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

a ta có a/b=c/d=>ac=bd.nhân cả 2 vế vs 3 ta được 3ac=3bd=>3a/b=3c/d

c từ ý a có 3a/b=3c/d=>3a/b+1=3c/d +1(cộng cả hai vế vs 1).sau đó quy đồng được 3a+b/b=3c+d/d

còn ý b thì hình như bạn chép sai r thì phải,đề bài đúng chắc là như thế nầy  a+b/b=c+a/a.nếu đề bài như thế thì sẽ giải giông ý c bạn nha!^^

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

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

\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)

\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)

\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)

Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)

a/ Ta có: ad = bc  => ac - ad  = ac - bc   => a . (a - d) = c . (a - b)  => \(\frac{a}{a-b}=\frac{c}{c-d}\)

b/ Ta có: ad = bc  => 3ac - ad = 3ac - bc  => a . (3c - d) = c . (3a - b)  => \(\frac{a}{3a-b}=\frac{c}{3c-d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

Ta có : 

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

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

Từ 1 và 2 

=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b.k,c=d.k\)

Ta có:

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

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

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