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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: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right).\left(b-6\right)=\left(a-5\right).\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow12a=10b\)
\(\Leftrightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(dpcm\right)\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow ab-6a+5b-30-ab-6a+5b+30=0\)
\(\Leftrightarrow-12a+10b=0\)
\(\Leftrightarrow-\frac{b}{12}=-\frac{a}{10}\)
\(\Leftrightarrow\frac{b}{12}=\frac{a}{10}\)
Tự lm tiếp
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\\ \Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\\ \Leftrightarrow12a=10b\\ \Leftrightarrow6a=5b\Leftrightarrow\dfrac{a}{b}=\dfrac{5}{6}\)
Bài 1 :
7^6+7^5-7^4=7^4.49+7^4.7-7^4.1
=7^4.(49+7-1)
=7^4.55
Vì 7^4.55 chia hết 5 Vậy 7^6+7^5-7^4 chia hết 5
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
(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
\(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Leftrightarrow ab-ab+5b+5b-30+30=6a+6a\)
\(\Leftrightarrow10b=12a\)
\(\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\left(đpcm\right)\)
- \(\frac{a+5}{a-5}\)=\(\frac{b+6}{b-6}\)(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
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
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 )