Đáp án:
Giải thích các bước giải:
Ta có: `\frac{ab+ac}{3}=\frac{ab+bc}{4}`
$⇔4(ab+ac)=3(ab+bc)$
$⇔4ab+4ac=3ab+3bc$
$⇔ab=3bc-4ac;4ac=3bc-ab$
Lại có: `\frac{ab+bc}{4}=\frac{bc+ac}{5}`
`⇔\frac{3bc-4ac+bc}{4}=\frac{bc+ac}{5}`
`⇔bc-ac=\frac{bc+ac}{5}`
$⇔5(bc-ac)=bc+ac$
$⇔5bc-bc=ac+5ac$
$⇔4bc=6ac$
$⇔2b=3a(1)$ (do $c>0$)
Lại có: `\frac{ab+bc}{4}=\frac{bc+ac}{5}`
`⇔\frac{ab+bc}{4}=\frac{4bc+4ac}{20}`
`⇔\frac{ab+bc}{4}=\frac{4bc+3bc-ab}{20}`
$⇔4(7bc-ab)=20(bc+ab)$
$⇔28bc-4ab=20bc+20ab$
$⇔28bc-20bc=20ab+4ab$
$⇔8bc=24ab$
$⇔c=3a(2)$ (do $b>0$)
Từ $(1);(2)⇒3a=2b=c(đpcm)$
