a: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

=>(a+5)(b-6)=(a-5)(b+6)

=>ab-6a+5b-30=ab+6a-5b-30

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

=>\(\dfrac{a}{b}=\dfrac{5}{6}\)

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

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

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

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

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

\(a>2,b>2\Rightarrow\left(a-2\right)\left(b-2\right)>0\)

\(\Rightarrow ab+4>2a+2b\)

\(\Rightarrow ab>a+b+\left(a+b-4\right)>a+b\)

\(\Rightarrow ab>a+b\) (đpcm)

\(a+b=ab=\dfrac{a}{b}\)

Ta có:

\(ab=\dfrac{a}{b}\Rightarrow ab=\dfrac{a^2}{ab}\)

\(\Rightarrow a^2b^2=a^2\)

\(\Rightarrow b^2=1\Rightarrow b=\pm1\)

Xét:

\(b=1\Rightarrow a+b=ab=\dfrac{a}{b}\Rightarrow a+1=a=a\left(KTM\right)\)

Xét:

\(b=-1\Rightarrow a+b=ab=\dfrac{a}{b}\Rightarrow a-1=-a=-a\)

\(\Rightarrow a-1=-a\)

\(\Rightarrow2a=1\Rightarrow a=\dfrac{1}{2}\)

Ta có:

\(\dfrac{a}{b}=a-1\rightarrowđpcm\)

\(b=-1\rightarrowđpcm\)

\(a=\dfrac{1}{2}\)

Vì: a>2 => a=2+m
b>2 => b=2+n (m, n thuộc N*)
=> a+b= (2+m) +(2+n)
a.b= (2+m). (2+n)
    = 2(2+n)+ m(2+n)
    = 4+ 2n+ 2m+ mn
    = 4+ m+ m+ n+ n+ mn
    = (4+ m+ n) +(m +n +mn)
    = (2+ m) +(2+ n) + (m+ n+ mn) > (2+ m)+ (2+n)
=> a.b > a+b .dpcm

Do b>2 => b>0

 mà a>2 => ab>2b (1)

 Tương tự ta có a>0 , b>2 => ab> 2a (2)

 cộng (1) vs (2) => ab+ab > 2a+2b

                         => 2ab > 2(a+b)

                         => ab > a+b (đpcm)

 ko tránh khỏi thiếu sót, nếu sai ai đó sửa lại nhé. Thắc mắc gì cứ hỏi

_Hết_

Ta có:\(a>2,b>2\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}< \dfrac{1}{2}+\dfrac{1}{2}=1\)

\(\Rightarrow\dfrac{b}{ab}+\dfrac{a}{ab}< 1\)

\(\Rightarrow\dfrac{a+b}{ab}< 1\)

\(\Rightarrow a+b< ab\left(đpcm\right)\)

theo đề bài ta có

`\(a>2,b>b\\ \Rightarrow\dfrac{1}{a}+\dfrac{1}{b}< \dfrac{1}{2}+\dfrac{1}{2}=1\\ \Rightarrow\dfrac{1}{a}+\dfrac{1}{b}< 1\\ \Rightarrow\dfrac{b}{ab}+\dfrac{a}{ab}< 1\left(do\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{b}{ab}+\dfrac{a}{ab}\right)\\ \Rightarrow\dfrac{a+b}{ab}< 1\\ \Rightarrow a+b< ab\left(đpcm\right)\)