Cho $\frac{1}{h}$ = $\frac{1}{2}$.($\frac{1}{a}$ + $\frac{1}{b}$) và $a$ $\neq$ $h$; $b$ $\neq$ $h$. Chứng minh rằng $\frac{a}{b}$ = $\frac{a-h}{h-b}$
Cho $\frac{1}{h}$ = $\frac{1}{2}$.($\frac{1}{a}$ + $\frac{1}{b}$) và $a$ $\neq$ $h$; $b$ $\neq$ $h$. Chứng minh rằng $\frac{a}{b}$ = $\frac{a-h}{h-b}$
$\quad \dfrac1h = \dfrac12\left(\dfrac1a + \dfrac1b\right)\qquad (a \ne h;\ b \ne h; a,\ b,\ h \ne 0)$
Ta có:
\(\begin{array}{l}
\quad \dfrac1h = \dfrac12\left(\dfrac1a + \dfrac1b\right)\\
\Leftrightarrow \dfrac2h = \dfrac1a + \dfrac1b\\
\Leftrightarrow \dfrac2h = \dfrac{a+b}{ab}\\
\Leftrightarrow 2ab = h(a + b)\\
\Leftrightarrow ab + ab = ah + bh\\
\Leftrightarrow ab – bh = ah – ab\\
\Leftrightarrow b(a – h) = a(h – b)\\
\Leftrightarrow \dfrac ab = \dfrac{a – h}{h-b}
\end{array}\)
Lời giải :
giải sử : `a/b = (a – h)/(h – b)`
`-> a (h – b) = b (a – h)`
`-> ah – ab = ab – bh`
`-> ah + bh = 2ab`
`-> h (a + b) = 2ab`
`-> a + b = (2ab)/h`
`-> (a + b)/(ab) = 1/h`
`-> 1/h = 1/2 . (a/(ab) + b/(ab) )`
`-> 1/h = 1/2 . (1/a + 1/b)`
Vậy `1/h = 1/2 (1/a + 1/b)` thì `a/b = (a – h)/(h – b)`