Question

find the x.
√2+√3/3√2-2√3​

Answer 1

Answer:

$\Large\boxed{\frac{12+5\sqrt{6}}{6}=4.04}$

Step-by-step explanation:

Given:

√2+√3/3√2-2√3​

To find the value of x, first you have to solve with square root from left to right numbers.

Solutions:

First, you have to multiply by conjugate from left to right numbers.

$\displaystyle \frac{3\sqrt{2}+2\sqrt{3}  }{3\sqrt{2}+2\sqrt{3}  }$

Rewrite the whole problem down.

$\displaystyle \frac{(\sqrt{2}+\sqrt{3}) \sqrt{3(3\sqrt{2}+2\sqrt{3})} }{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}$

Solve.

Used foil method.

$\Large\boxed{\textnormal{FOIL METHOD FORMULA}}$

$\displaystyle (A+B)(C+D)=AC+AD+BC+BD$

A=√2

B=√3

C=3√2

D=2√3

Rewrite the problem.

$\displaystyle \sqrt{2}*3\sqrt{2}+\sqrt{2}*2\sqrt{3}+\sqrt{3}*3\sqrt{2}+\sqrt{3}*2\sqrt{3}$

$\displaystyle 3\sqrt{2}\sqrt{2}+2\sqrt{2}\sqrt{3}+3\sqrt{3}\sqrt{2}+2\sqrt{3}\sqrt{3}$

Solve.

$\displaystyle 3\sqrt{2}\sqrt{2}+2\sqrt{2}\sqrt{3}+3\sqrt{3}\sqrt{2}+2\sqrt{3}\sqrt{3}=12+5\sqrt{6}$

Used two squares formula.

$\Large\boxed{\textnormal{TWO SQUARES FORMULA}}$

$\displaystyle (A-B)(A+B)=A^2-B^2$

A=3√2

B=2√3

$\displaystyle \left(3\sqrt{2}\right)^2-\left(2\sqrt{3}\right)^2$

Solve. (Simplify/and refine.)

$\left(3\sqrt{2}\right)^2-\left(2\sqrt{3}\right)^2=\boxed{6}$

$\large\displaystyle \boxed{\frac{12+5\sqrt{6}}{6}=4.04}}$

The correct answer is 12+5√6/6=4.04.