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

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

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

=> 10b = 12a

=> 5b = 6a

=> \(\frac{a}{b}=\frac{5}{6}\)(đpcm)

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

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

<=> \(1+\frac{10}{a-5}=1+\frac{12}{b-6}\)

<=> \(\frac{10}{a-5}=\frac{12}{b-6}\)

<=> 10( b - 6 ) = 12( a - 5 )

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

<=> 5b - 30 = 6a - 30

<=> 5b = 6a

<=> 5/6 = a/b ( đpcm )

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\\ \Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\\ \Leftrightarrow12a=10b\\ \Leftrightarrow6a=5b\Leftrightarrow\dfrac{a}{b}=\dfrac{5}{6}\)

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

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

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

 (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 : 

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

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

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

nhân ra ik ròi suy ra đpcm :D

Cho a là số tự nhiênchia 6 dư 2 và b là số tự nhiên chia 6 dư 3. Chứng minh axb chia hết cho 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

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

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

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

\(\Leftrightarrow10b=12a\)

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

• \(\frac{a+5}{a-5}\)=\(\frac{b+6}{b-6}\)(ta hoán đổi trung tỉ)=>\(\frac{a+5}{b+6}\)=\(\frac{a-5}{b-6}\)=>\(\frac{\left(a+5\right)-5}{\left(b+6\right)-6}\)=\(\frac{\left(a+5\right)-a}{\left(b+6\right)-b}\)=a/b=5/6

1) a chia 6 dư 2 => a= 6k+2

b chia 6 dư 3 => b= 6k+3

=> ab=\(\left(6k+2\right)\left(6k+3\right)=36k^2+30k+6\)=> chia hết cho 6 

2) a= 5k+2; b=5k+3

=> \(ab=\left(5k+2\right)\left(5k+3\right)=25k^2+25k+6=25k\left(k+1\right)+6\)

=> dễ thấy 25k(k+1) chia hết cho 5. 6 chia 5 dư 1

=> ab chia 5 dư 1