Question

Stiind ca a+b=18si a×b=72, calculeaza 1supra a+1 supra b

Answer 1

$a + b = 18$

$a \times b = 72$

$a = \frac{72}{b}$

$\frac{72}{b} + b = 18 \: | \times b$

$72 + {b}^{2} = 18b$

${b}^{2} - 18b + 72 = 0$

$x = 1$

$y = - 18$

$z = 72$

$delta = {y}^{2} - 4xz$

$delta = ( { - 18})^{2} - 4 \times 1 \times 72$

$delta = 324 - 288$

$delta = 36$

$b(1,2) = \frac{ - y + - \sqrt{delta} }{2x}$

$b(1,2) = \frac{ - ( - 18) + - \sqrt{36} }{2 \times 1}$

$b(1,2) = \frac{18 + - 6}{2}$

$b(1) = \frac{18 + 6}{2}$

$b(1) = \frac{24}{2}$

$b(1) = 12$

$b(2) = \frac{18 - 6}{2}$

$b(2) = \frac{12}{2}$

$b(2) = 6$

$a + b = 18$

$a + 12 = 18$

$a(1) = 18 - 12$

$a(1) = 6$

$a + 6 = 18$

$a(2) = 18 - 6$

$a(2) = 12$

$\frac{1}{a} + \frac{1}{b}$

$= \frac{1}{6} + \frac{1}{12}$

$= \frac{2}{12} + \frac{1}{12}$

$= \frac{2 + 1}{12}$

$= \frac{3}{12}$

$= \frac{1}{4}$

$\frac{1}{a} + \frac{1}{b}$

$= \frac{1}{12} + \frac{1}{6}$

$= \frac{1}{12} + \frac{2}{12}$

$= \frac{1 + 2}{12}$

$= \frac{3}{12}$

$= \frac{1}{4}$