Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)

\(\Rightarrow\frac{5a+5b}{5b}=\frac{5b\left(k+1\right)}{5b}=k+1\)

\(\frac{c^2+cd}{cd}=\frac{k^2d^2+kd^2}{kd^2}=\frac{kd^2\left(k+1\right)}{kd^2}=k+1\)

\(\Rightarrow\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)
 

\(\)\(\frac{5a+5b}{5b}=\frac{5a}{5b}+\frac{5b}{5b}=\frac{a}{b}+1\)

\(\frac{c^2+cd}{cd}=\frac{c^2}{cd}+\frac{cd}{cd}=\frac{c}{d}+1\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)

\(\Rightarrowđpcm\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a= b.k ; c= d.k

\(\frac{5a+5b}{5b}=\frac{5a}{5b}+\frac{5b}{5b}=\frac{a}{b}+1=\frac{b.k}{b}+1=k+1\left(1\right)\)

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

Từ \(\left(1\right);\left(2\right)\Rightarrowđpcm\)

a)a/b=c/d=a+b/c+d=a-b/c-d(tc day ti so bang nhau)

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

b)a/b=c/d=>5a/5b=2c/2d=5a+2c/5c+2d(*) va a/b=4c/4d=a-4c/c-4d(**)

c)a/b=c/d=a+b/c+d=>(a/b)^2=ab/cd=(a+b/c+d)^2

đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

a)

\(\frac{5a+2c}{5b+2d}=\frac{5bk+2dk}{5b+2d}=\frac{k\left(5b+2d\right)}{5b+2d}=k\)

\(\frac{a-4c}{b-4d}=\frac{bk-4dk}{b-4d}=\frac{k\left(b-4d\right)}{b-4d}=k\)

=>\(\frac{5a+2c}{5b+2d}=\frac{a-4c}{b-4d}=k\)(đpcm)

b)

\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}=\frac{b}{d}\)

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

=>\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)

\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)

Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)