Question

8
-
Rezolvați, în R, ecuațiile:
25
a) x2 = 5;
d) 12x² +7==;
3
e) x² + 3 =(-2);
2
3
c) 2-r? = 0;
f) x² + 3 =1,(6).
2 2
b) 12
Care ma ajuta cu ex 8 ii dau coroana nu corona

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Răspuns: Ai rezolvarea mai jos

Explicație pas cu pas:

Salutare!

$\bf\:(a)\:\:x^{2} = 5 \implies x_{1} = -\sqrt{5} \:\:si \:\:x_{2} = \sqrt{5}$

$\bf\:(b)\:\:t^{2} = \dfrac{1}{2} =\sqrt{\dfrac{1}{2}} =\dfrac{1}{\sqrt{2}} \implies t_{1} = -\dfrac{\sqrt{2}}{2} \:\:si \:\:t_{2} =\dfrac{\sqrt{2}}{2}$

$\bf\:(c)\:\:2\cdot x^{2} = 0 \implies x^{2} = 0\implies x = 0$

$\bf\:(d)\:\:12x^{2}+7 = \dfrac{25}{3}\:\:\:\Big|\cdot 3$

$\bf 36x^{2}+21 = 25$

$\bf 36x^{2} = 25 -21$

$\bf 36x^{2} = 4\:\:\:\Big|:4$

$\bf 9x^{2} =1\implies x^{2} = \dfrac{1}{9}\implies x_{1} = \sqrt{\dfrac{1}{9}}= \dfrac{1}{3}\:\:\:si \:\:\:x_{2} = -\dfrac{1}{3}$

$\bf\:(e)\:\:x^{2}+3^{3} = (-2)^{4}$

$\bf x^{2}+27 = 2^{4}$

$\bf x^{2}+27 = 16$

$\bf x^{2} = 16 -27$

$\bf x^{2} = - 11 \implies x \in \not0$

$\bf\:(f)\:\:\dfrac{1}{2} x^{2}+\dfrac{3}{2} = \dfrac{16-1}{9}$

$\bf \dfrac{1}{2} x^{2}+\dfrac{3}{2} = \dfrac{5}{3}\:\:\:\Big|\cdot 6$

$\bf 3x^{2}+9 = 10$

$\bf 3x^{2} = 10-9$

$\bf 3x^{2} = 1$

$\bf x^{2} = \dfrac{1}{3}\implies x_{1}=-\dfrac{\sqrt{3} }{3}\:\:\:si\:\:\: x_{2}=\dfrac{\sqrt{3} }{3}$

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