Question

Determinati nr real ''x'' care verifica egalitatile:
a)x supra 2√5=√15 supra 10
b)√6 supra x=√2 supra √6
c)x supra √2 - 3 supra √6=√2 supra √3
d)x supra √30 - √27 supra √10=2√30 supra 10
e)x(√5-√3)=5 supra √5 - 3+√3 supra √3 + 1

Efectuati urmatoarele calcule,amplificand formula(a+b)(a-b)=a la puerea 2-b la puterea 2:
a) (√6+√2)(√6-√2)=
b) (√5+√3)(√5-√3)=
c) (√18-√11)(√18+√11)=
d) (2+√3)(2-√3)=
e) (5-√2)(5+√2)=
f) (√7-1)(√7+1)=
g) (2√3-3)(2√3+3)=
h) (7+4√3)(7+4√3)=
i) (5√3+6√2)(5√3-6√2)=
j) (6√2-8)(6√2+8)=
k) (5√3+6√2)(3√5+2√6)=
l) (3√5+2√11)(3√5-2√11)=

Calculati suma inverselor numerelor a,b,c,unde:
a) a=√3+√2,b=√4+√3,c=√5+√4
b) a=√5+2,b=2√5+4,c=3+√5
c) a=3√2-2√3,b=3-2√2,c=3√3-5.
Multumesc daca se poate un raspuns rapid,va rog mult ajutati-ma:3

Answer 1

a)x supra 2√5=√15 supra 10
b)√6 supra x=√2 supra √6
c)x supra √2 - 3 supra √6=√2 supra √3
d)x supra √30 - √27 supra √10=2√30 supra 10
e)x(√5-√3)=5 supra √5 - 3+√3 supra √3 + 1
Astea 5 vrei.
$\frac{x}{2\sqrt{5}}=\frac{\sqrt{15}}{10}$
$10x=10\sqrt{3}\Rightarrow x=\sqrt{3}$
b)$\frac{\sqrt{6}}{x}=\frac{\sqrt{2}}{\sqrt{6}}$
$</span>\sqrt{2}x=6 \Rightarrow x=\frac{6\sqrt{2}}{2}$
c)$\frac{x}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{3}}$
$\sqrt{3}(x-\sqrt{3})=2\Leftrightarrow x\sqrt{3}-3=2 \Leftrightarrow x=\frac{5}{\sqrt{3}}$
d)$\frac{x}{\sqrt{30}}-\frac{\sqrt{27}}{\sqrt{10}}=\frac{2\sqrt{30}}{10}$
$\frac{x\sqrt{10}-9\sqrt{10}}{10\sqrt{3}}=\frac{6\sqrt{10}}{10\sqrt{3}}$
De unde x=15
e)$x(\sqrt{5}-\sqrt{3})=\sqrt{5}-\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}$
$</span><span>x(\sqrt{5}-\sqrt{3})=</span>\sqrt{3}-\sqrt{3}$
De unde x=1