Question

Daca $\frac{a}{b}$=|1-$\sqrt{2}|$ sa se afle $\frac{a^{2}+b^{2} }{ab}$


Ajutati va rog, dau coroana

Answer 1

Răspuns:

2√2

Explicație pas cu pas:

a/b = I 1-√2 I

√2 > 1  => I 1-√2 I = √2 - 1

a/b = √2 - 1  => a = b·(√2-1)

a² = (b·√2-1)² = b²·(2+1-2√2) = b²·(3-2√2)

(a²+b²)/ab = (b²·(3-2√2) + b²) / [ b·(√2-1)·b] =

= b²·(3-2√2+1) / [b²·(√2-1)] =

= (4-2√2) / (√2-1) = (4-2√2)(√2+1) / [(√2-1)(√2+1)] =

= 4√2+4-4-2√2 = 2√2

Answer 2

$\it \dfrac{a}{b}=|\underbrace{1-\sqrt2}_{<0}| \Rightarrow \dfrac{a}{b}=\sqrt2-1\ \ \ \ \ (*)\\ \\ \\ \dfrac{a^2+b^2}{ab}=\dfrac{a^2}{ab}+\dfrac{b^2}{ab}=\dfrac{a}{b}+\dfrac{b}{a}\ \stackrel{(*)}{=}\ \sqrt2-1+\dfrac{^{\sqrt2+1)}1}{\ \ \sqrt2-1}=\sqrt2-1+\sqrt2+1=2\sqrt2$