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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
(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 :
\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{\left(b+6\right)-\left(b-6\right)}{\left(a+5\right)-\left(a-5\right)}=\frac{12}{10}=\frac{6}{5}\)
\(\Rightarrow5\left(b-6\right)=6\left(a-5\right)\Leftrightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow ab-ab+5b+5b-30+30=6a+6a\)
\(\Leftrightarrow10b=12a\)
\(\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)
- \(\frac{a+5}{a-5}\)=\(\frac{b+6}{b-6}\)(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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow5b=6a\)
\(\Rightarrow\frac{a}{b}=\frac{5}{6}\)
Đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
<=> \(\frac{a-5+10}{a-5}=\frac{b-6+12}{b-6}\)
<=> \(1+\frac{10}{a-5}=1+\frac{12}{b-6}\)
<=> \(\frac{10}{a-5}=\frac{12}{b-6}\)
<=> 10( b - 6 ) = 12( a - 5 )
<=> 5( b - 6 ) = 6( a - 5 )
<=> 5b - 30 = 6a - 30
<=> 5b = 6a
<=> \(\frac{6}{5}=\frac{b}{a}\)hay \(\frac{a}{b}=\frac{5}{6}\)( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{x+5}{x-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow16a=10b\)
\(\Leftrightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\frac{a-5+10}{a-5}=\frac{b-6+12}{b-6}\)
\(\Leftrightarrow1+\frac{10}{a-5}=1+\frac{12}{b-6}\)
\(\Leftrightarrow\frac{10}{a-5}=\frac{12}{b-6}\)
\(\Rightarrow10.\left(b-6\right)=12.\left(a-5\right)\)
\(10b-60=12a-60\)
\(10b=12a\)
\(\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(=>\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{a+5-\left(a-5\right)}{b+6-\left(b-6\right)}=\frac{a+5-a+5}{b+6-b+6}=\frac{10}{12}=\frac{5}{6}\)
\(\frac{a+5}{b+6}=\frac{a-5}{b-6}=\frac{a+5+\left(a-5\right)}{b+6+\left(b-5\right)}=\frac{a+5+a-5}{b+6+b-6}=\frac{2a}{2b}=\frac{a}{b}\)
=>\(\frac{a}{b}=\frac{a+5}{b+5}=\frac{5}{6}\)
=>\(\frac{a}{b}=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Áp dụng tính chất của tỉ lệ thức ta có:\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Mặt khác: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)
Cho \(\frac{a+5}{a-5}=\frac{b+6}{b-6}\) (a khác 5, b khác 6). Chứng minh \(\frac{a}{b}=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
Ta có : \(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\frac{b-6}{a-5}=\frac{b+6}{a+5}\)
Áp dụng t/c dãy tỉ số bằng nhau :
\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{\left(b+6\right)-\left(b-6\right)}{\left(a+5\right)-\left(a-5\right)}=\frac{12}{10}=\frac{6}{5}\)
\(\Rightarrow5\left(b-6\right)=6\left(a-5\right)\Leftrightarrow5b-30=6a-30\Leftrightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(ab-6a+5b-30=ab+6a-5b-30\)
\(5b-6a=6a-5b\)
\(6a=5b\Rightarrow\frac{a}{5}=\frac{b}{6}\)