Cho 1/c=1/2 (1/a+1/b) ( vs a,b,c ≠0, b≠c). CMR a/b=a-c/c-b Giúp mk vs, hết điểm r 30/11/2021 Bởi Alice Cho 1/c=1/2 (1/a+1/b) ( vs a,b,c ≠0, b≠c). CMR a/b=a-c/c-b Giúp mk vs, hết điểm r
`1/c=1/2(1/a+1/b)` `⇒2/c=1/a+1/b=(a+b)/(ab)` `⇒2ab=(a+b)c=ac+bc` `⇒ab-ac=bc-ab` `⇒a(b-c)=b(c-a)` `⇒a/b=(c-a)/(b-c)=(a-c)/(c-b)` Bình luận
$\dfrac1c =\dfrac12\left(\dfrac1a +\dfrac1b\right)$ $\to \dfrac1c =\dfrac{a+b}{2ab}$ $\to 2ab = c(a+b)$ $\to ab – ac = bc – ab$ $\to a(b-c) = b(c-a)$ $\to \dfrac ab =\dfrac{c-a}{b-c}$ $\to \dfrac ab =\dfrac{a – c}{c -b}$ (đpcm) Bình luận
`1/c=1/2(1/a+1/b)`
`⇒2/c=1/a+1/b=(a+b)/(ab)`
`⇒2ab=(a+b)c=ac+bc`
`⇒ab-ac=bc-ab`
`⇒a(b-c)=b(c-a)`
`⇒a/b=(c-a)/(b-c)=(a-c)/(c-b)`
$\dfrac1c =\dfrac12\left(\dfrac1a +\dfrac1b\right)$
$\to \dfrac1c =\dfrac{a+b}{2ab}$
$\to 2ab = c(a+b)$
$\to ab – ac = bc – ab$
$\to a(b-c) = b(c-a)$
$\to \dfrac ab =\dfrac{c-a}{b-c}$
$\to \dfrac ab =\dfrac{a – c}{c -b}$ (đpcm)