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[tex]\bf 5. \\ \\ \it \ \ \Rightarrow a=\dfrac{1}{\sqrt3}(1-1+\dfrac{1}{3})=\dfrac{1}{3\sqrt3}\\ \\ \\ \Rightarrow b=\dfrac{\sqrt6}{\sqrt8}(1+\dfrac{3}{2})=\dfrac{\sqrt3}{2}\cdot\dfrac{5}{2}=\dfrac{5\sqrt3}{4}\\ \\ \\ a\cdot b=\dfrac{1}{3\sqrt3}\cdot\dfrac{5\sqrt3}{4}=\dfrac{5}{12}[/tex]

7.

[tex]\it x\sqrt{27}-\sqrt{12}=\sqrt{300}|_{:\sqrt3}\ \Rightarrow 3x-2=10|_{+2} \Rightarrow 3x=12|_{:3} \Rightarrow x=4[/tex]

Răspuns:

Explicație pas cu pas:

5.

a = 1/√3 - 2/2√3 + 1/3√3 = 1/√3 - 1/√3 + 1/3√3 = 1/3√3

b = √6/√8 + √54/√32 = √3/2 + 3√6/4√2 = √3/2 + 3√3/4 = 2√3/4 + 3√3/4

= 5√3/4

a*b = 1/3√3 *5√3/4 = 5/12

_______________

7.

x√27 - √12 = √300

3x√3 - 2√3 = 10√3

√3*(3x - 2) = 10√3

3x - 2 = 10

3x = 10 + 2 = 12

x = 12 : 3 = 4

_________________

8.

√4*y + √72 = √128

2√y + 6√2 = 8√2

2√y = 8√2 - 6√2 = 2√2

y = 2