Question

Îmi trebuie urgent !!!
Subiectul al ll-lea ex. 2, 3 și 4
Va rog!! Ofer 30 de pct

Answer 1

Răspuns:

Explicație pas cu pas:

ex2.

$a=\frac{2}{\sqrt{2}}+1=\frac{2\sqrt{2}}{2}+1=\sqrt{2}+1\\b=\sqrt{(1-\sqrt{2})^{2}}=|1-\sqrt{2}|=\sqrt{2}-1\\m_{a}=(a+b)/2=(\sqrt{2}+1+\sqrt{2}-1):2=\frac{2\sqrt{2}}{2}=\sqrt{2}\\m_{g}=\sqrt{a*b}=\sqrt{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{(\sqrt{2})^{2}-1^{2}=\sqrt{2-1}=1$

ex3

AB=$\sqrt{(4-(-2))^{2}+(-4-5)^{2}}=\sqrt{36+81}=\sqrt{117}=\sqrt{9*13} =3\sqrt{13}$

ex4.

Exspresia E este trinom de gradul II, valoarea minim[ o ia ]n virful parabolei

$x_{V}=\frac{-b}{2a}=\frac{-4}{2*4}=-\frac{1}{2}\\E_{min}=E(x_{V})=E(-\frac{1}{2})=4*(-\frac{1}{2})^{2}+4*(-\frac{1}{2})+11=1-2+11=10$

Answer 2

$\it2)\ a=\dfrac{2}{\sqrt2}+1=\dfrac{\sqrt2\cdot\sqrt2}{\sqrt2}+1 =\sqrt2+1\\ \\ b=\sqrt{(1-\sqrt2)^2} =|1-\sqrt2| =-1+\sqrt2=\sqrt2-1\\ \\ m_a=\dfrac{a+b}{2}=\dfrac{\sqrt2+1+\sqrt2-1}{2} =\dfrac{2\sqrt2}{2} =\sqrt2\\ \\ m_g=\sqrt{a\cdot b} =\sqrt{(\sqrt2-1)(\sqrt2+1)}=\sqrt{2-1} =\sqrt1=1$

$\it 3)\ AB^2=(x_B-x_A)^2+(y_B-y_A)^2=(4+2)^2+(-4-5)^2 =36+81=117\\ \\ \\ AB=\sqrt{117}=\sqrt{9\cdot13} =3\sqrt{13}$

$\it E(x)=4x^2+4x+11=4x^2+4x+1+10=(2x+1)^2+10 \geq 10 \Rightarrow\\ \\ \Rightarrow min\ E(x)=10,\ pentru\ 2x+1=0 \Rightarrow x=-\dfrac{1}{2}$