Ta có : \(\frac{a+5}{b-5}\frac{b+6}{b-6}\)

=> \(\frac{a+5}{b+6}=\frac{a-5}{b-6}\)

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

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

\(\frac{2a}{2b}=\frac{5.2}{6.2}=\frac{10}{12}\)

=> \(\frac{a}{b}=\frac{5}{6}\left(\text{đ}pcm\right)\)

Từ

\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\frac{b-6}{a-5}=\frac{b+6}{a-5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau . Ta có

\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{b-6+b+6}{a-5+a+5}=\frac{2b}{2b}=\frac{b}{a}=\frac{b+6-b}{a+5-a}=\frac{6}{5}\)

\(\Rightarrow\frac{a}{b}=\frac{5}{6}\) (đpcm)

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\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)

\(\Rightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)

 (Đpcm)

Từ \(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\frac{b-6}{a-5}=\frac{b+6}{a+5}\)

Áp dụng t/c dãy tỉ số bằng nhau : 

\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{\left(b+6\right)-\left(b-6\right)}{\left(a+5\right)-\left(a-5\right)}=\frac{12}{10}=\frac{6}{5}\)

\(\Rightarrow5\left(b-6\right)=6\left(a-5\right)\Leftrightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)

ta có : \(\frac{a+5}{a-5}\)=\(\frac{b+6}{b-6}\)

          <=> (a+5)(b-6)=(a-5)(b+6)

          <=> ab-6a+5b-30=ab+6a-5b-30

         <=>5b+5b=6a+6a

           <=> 10b=12a

          <=> \(\frac{a}{b}\)=\(\frac{10}{12}\)=\(\frac{5}{6}\)=> đfcm

A-5+10/A-5=

A-5/A-5

NGU

bạn nhân chéo rồi rút gọn nha

sau đó áp dụng tính chất cuartir lệ thức

\(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)

\(\Leftrightarrow ab-ab+5b+5b-30+30=6a+6a\)

\(\Leftrightarrow10b=12a\)

\(\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)

a+5/a-5=b+6/b-6=a+5/b+6=b+6/b-69 (giao hoán trung tỉ)=>(a+5)-5/(b+6)-6=(a+5)-a/(b+6)-6=a/b=5/6

BẠN ƠI CÁI MÀ '/' LÀ PHẦN NHA

Vì \(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\frac{a+5}{b+6}=\frac{a-5}{b-6}\)

Áp dụng TC DTSBN ta có :

\(\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{\left(a+5\right)+\left(a-5\right)}{\left(b+c\right)+\left(b-6\right)}=\frac{2a}{2b}=\frac{a}{b}\) (1)

\(\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{\left(a+5\right)-\left(a-5\right)}{\left(b+6\right)-\left(b-6\right)}=\frac{10}{12}=\frac{5}{6}\)(2)

Từ (1) ; (2) \(\Rightarrow\frac{a}{b}=\frac{5}{6}\)(đpcm)

\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\Rightarrow a.b+5.b-6a-30=a.b-5.b+6.a-30\Rightarrow a.b+10.b=a.b+12a\)\(\Rightarrow10.b=12.a\Rightarrow5.b=6.a\Rightarrow\frac{a}{b}=\frac{5}{6}\left(đpcm\right)\)

\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

                                     =>   6a = 5b  

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

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

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

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

=> 10b = 12a

=> \(\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\)(đpcm)

\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\) => (a + 5)(b - 6) = (a - 5)(b + 6) => ab - 6a + 5b - 30 = ab + 6a - 5b - 30 

=>\(-6a+5b\) = 6a - 5b =\(-\left(6a-5b\right)\)  => 6a - 5b = 0 => 6a = 5b => \(\frac{a}{b}=\frac{5}{6}\)(đpcm)

\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{2a}{2b}=\frac{a}{b}\)

Suy ra:\(\frac{a+5}{b+6}=\frac{a}{b}\Rightarrow a.\left(b+6\right)=b.\left(a+5\right)\)

=>ab+6a=ab+5b

=>6a=5b

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

Phan Thanh Tịnh lớp 7 có lẽ ko có khả năng làm như vậy