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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 :
\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{\left(b+6\right)-\left(b-6\right)}{\left(a+5\right)-\left(a-5\right)}=\frac{12}{10}=\frac{6}{5}\)
\(\Rightarrow5\left(b-6\right)=6\left(a-5\right)\Leftrightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow5b=6a\)
\(\Rightarrow\frac{a}{b}=\frac{5}{6}\)
Đpcm
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Leftrightarrow\left(a+5\right).\left(b-6\right)=\left(a-5\right).\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow-6a+5b=6a-5b\Rightarrow-6a+10b=6a\Rightarrow10b=12a\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)
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
\(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow ab-ab+5b+5b-30+30=6a+6a\)
\(\Leftrightarrow10b=12a\)
\(\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)
- \(\frac{a+5}{a-5}\)=\(\frac{b+6}{b-6}\)(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
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
<=> \(\frac{a-5+10}{a-5}=\frac{b-6+12}{b-6}\)
<=> \(1+\frac{10}{a-5}=1+\frac{12}{b-6}\)
<=> \(\frac{10}{a-5}=\frac{12}{b-6}\)
<=> 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 )
Bài 1:
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^4=0\)
=>2x(2x-1)(2x-2)=0
hay \(x\in\left\{0;\dfrac{1}{2};1\right\}\)
Bài 3:
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Leftrightarrow\dfrac{a-5+10}{a-5}=\dfrac{b-6+12}{b-6}\)
\(\Leftrightarrow\dfrac{10}{a-5}=\dfrac{12}{b-6}\)
\(\Leftrightarrow\dfrac{a-5}{5}=\dfrac{b-6}{6}\)
\(\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{6}\)
hay a/b=5/6
Ta có: \(\frac{x+5}{x-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow16a=10b\)
\(\Leftrightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)
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 :
\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{\left(b+6\right)-\left(b-6\right)}{\left(a+5\right)-\left(a-5\right)}=\frac{12}{10}=\frac{6}{5}\)
\(\Rightarrow5\left(b-6\right)=6\left(a-5\right)\Leftrightarrow5b-30=6a-30\Leftrightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(ab-6a+5b-30=ab+6a-5b-30\)
\(5b-6a=6a-5b\)
\(6a=5b\Rightarrow\frac{a}{5}=\frac{b}{6}\)