Question

ex5,6 va rog !-Dau coroana !​

Answer 1

5)

a = x² - 4x + 5 ⇒ a = x - 4x + 4 + 1 ⇒ a = (x-2)² + 1

Dar, (x-2)² ≥ 0 ⇒ (x-2)² + 1 ≥ 1  ⇒ a ≥ 1  ⇒ a pozitiv.

6)

$b = \sqrt{5-2\sqrt 6}+\sqrt{5+2\sqrt 6}-\sqrt{(4+2\sqrt 3)^2} \\ \\ b =\sqrt{(\sqrt 2-\sqrt 3)^2}+\sqrt{(\sqrt 2+\sqrt 3)^2}-\sqrt{(4+2\sqrt 3)^2}\\ \\ b =|\sqrt 2-\sqrt 3| + |\sqrt 2+\sqrt 3|-|4+2\sqrt 3| \\ \\ b =(\sqrt 3-\sqrt 2)+\sqrt 2+\sqrt 3 -(4+2\sqrt 3) \\ \\ b =2\sqrt 3 - 4 - 2\sqrt 3 \\ \\ \Rightarrow \boxed{b = -4 \in \mathbb{Z}}$

Answer 2

Răspuns:

Explicație pas cu pas:

ex 5/ a este reprezentat ca o funcţie de gradul II şi ea va fi strict pozitivă dacă a=1>0 si Δ<0. Δ=(-4)²-4·1·5=16-20=-4<0, deci a<0 pt. orice x∈R.

ex 6.  

$b=\sqrt{5-2\sqrt{6}}+ \sqrt{5+2\sqrt{6}}-\sqrt{(4+2\sqrt{3})^{2}} =\\=\sqrt{(\sqrt{3})^{2} -2*\sqrt{3} *\sqrt{2}+(\sqrt{2})^{2}}+ \sqrt{(\sqrt{3})^{2} +2*\sqrt{3} *\sqrt{2}+(\sqrt{2})^{2}}-|4+2\sqrt{3}|=\sqrt{(\sqrt{3}-\sqrt{2})^{2}}+ \sqrt{(\sqrt{3}+\sqrt{2})^{2}}-(4+2\sqrt{3})=|\sqrt{3}-\sqrt{2}|+|\sqrt{3}+\sqrt{2}| -4-2\sqrt{3}=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-4-2\sqrt{3}=2\sqrt{3}-4-2\sqrt{3}=-4.$

-4∈Z, deci  b∈Z