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

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 \(\frac{a}{3a+b}=\frac{b}{3c+d}\)

a/b=c/d => a/c=b/d
            Mà a/c=3a/3c
=> 3a/3c=b/d
Áp dụng t/c của dãy tỉ số bằng nhau , ta có :
  3a/3c=b/d = 3a+b/ 3c+d
Ta có a/c=3a+b/3c+d
   => a/3a+b=c/3c+4
 

vi a/b=c/d =>a/c=b/d ma a/c3a/3c=>b/d=3a/3c

ap dung t/c day ti so bang nhau ta co :

3a/3c=b/d/=3a+b/3c/d

ta co a/c =3a+b/3c+d

=>a/3a+b=c/3c+d

cách 1:

ta có:

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

\(\Rightarrow a=b\cdot k;c=d\cdot k\left(1\right)\)

ta có:

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

thay (1) vào a ta được:

\(\frac{b\cdot k}{3b\cdot k+b}=\frac{b\cdot k}{b\cdot\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)

thay (1) vào c ta được:

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

từ (1) và (2)\(\Rightarrowđpcm\)

cách 2:

ta có:
\(\frac{a}{b}=\frac{c}{d}\)
=>ad=bc
=>3ac+ad=3ac+bc3ac+ad=3ac+bc

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

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

a, Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}\)

Áp dụng tính chất của day tỉ số bằng nhau ta được:

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

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

Bài 1:

Ta có:\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta được:

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

(ĐPCM)

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

Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=x\)

Xét \(x^2=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)

=>(đpcm)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

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

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

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

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

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

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

=> \(\frac{3a}{3c}=\frac{5b}{5d}\)

Áp dụng dãy tỉ số bằng nhau ta có:

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

=> Đpcm

Chúc bạn làm bài tốt

vì a/b= c/d

⇒ a+b/c+d=3a+5b/3c+5d=3a-5b/3c-5d

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

3a+5b/3c+5d=3a-5b/3c-5d

⇒ 3a+5b/3a-5b=3c+5d/3c-5d (đpcm)

đúng 100 %

Chúc bạn học tốt

Theo dãy tỉ số (=) ta* có: 

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

=> a = 1/3 . 3b = b (1)

=> b = 1/3 . 3c =  c (2)

=> c = 1/3 . 3d = d (3)

Từ(1) (2) và (3) =. a = b= c =d => ĐPCM