Question

a.\(\sqrt{2}.x-\sqrt{50}=0\)

b.\(\sqrt{3}.x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)

c.\(\sqrt{3}.x^2-\sqrt{12}=0\)

d.\(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\)

Answer 1

a) \(\sqrt{2}\cdot x-\sqrt{50}=0< =>\sqrt{2}\cdot x=\sqrt{50}\)

<=> x= 5

b) \(\sqrt{3}\cdot x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)

<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot\sqrt{4}+\sqrt{3}\cdot\sqrt{9}\)

<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot5< =>x+1=5\)

<=> x=4

Answer 2

c) \(\sqrt{3}\cdot x^2-\sqrt{12}=0\\ < =>x^2=\sqrt{4}=2;-2\\ < =>x=\sqrt{2};-\sqrt{2}\)

d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\\ < =>x^2=\sqrt{100}=10;-10\\ < =>x=\sqrt{10};-\sqrt{10}\)