+0

+1
590
5

{2-√3} / {2+√3}

Feb 26, 2018
edited by ManuelBautista2019  Feb 26, 2018
edited by ManuelBautista2019  Feb 26, 2018
edited by ManuelBautista2019  Feb 26, 2018

#1
+1

7-4√3

8)

Feb 26, 2018
#2
+1

$$2-\frac{\sqrt3}{2}+\sqrt3\\ =2\qquad +\sqrt3-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3}{2}-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3-1\sqrt3}{2}\\ =2\qquad +\frac{1\sqrt3}{2}\\ =2+\frac{\sqrt3}{2}\\$$

.
Feb 26, 2018
#3
0

(2-3^(1/2))/(2+3^(1/2)) or 2-3^1/2

2+3^1/2

3^1/2 cancels and you are left with 2/2 which equals one.

^ is the same thing as an exponent.

and to clarify 3^1/2, that is the same as saying the square root of 3.

Feb 27, 2018
#4
+1

It appears as if this question has been edited since the replies of other answerers. As of now, the expression to simplify is $$\frac{2-\sqrt{3}}{2+\sqrt{3}}$$

 $$\frac{2-\sqrt{3}}{2+\sqrt{3}}$$ Radicals in the denominator are considered to be improper; it is required that all radical terms are moved to the numerator. This can be achieved by utilizing the conjugate of the denominator: $$2-\sqrt{3}$$. $$\frac{2-\sqrt{3}}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}$$ To simplify the denominator, it is possible to use the algebraic rule that $$(a+b)(a-b)=a^2-b^2$$. Notice that this process does not change the inherent value of the expression since all I am doing is multiplying by 1, which has no effect. $$\frac{\left(2-\sqrt{3}\right)^2}{2^2-\sqrt{3}^2}$$ By squaring the radical in the denominator, we are getting rid of a radical's existence from the denominator. $$\frac{\left(2-\sqrt{3}\right)^2}{4-3}$$ Well, it is very clear at this point that the denominator simplifies to 1. Expand the numerator using the rule that $$(a-b)^2=a^2-2ab+b^2$$ $$2^2-2*2*\sqrt{3}+\sqrt{3}^2$$ Simplify. $$4-4\sqrt{3}+3$$ $$7-4\sqrt{3}$$ This is what OfficialBubbleTanks got.
Feb 27, 2018
#5
+1

Thanks To all of you!!! Feb 27, 2018