Question

If √(2 )=1.4142,√3=1.732 , find the value of 4/(3√2-2√3)+3/(3√3+2√2)

Answer 1

Given,

$\sqrt{2} = 1.4142 \\ \sqrt{3 } = 1.732$

To find :-

$\frac{4}{(3 \sqrt{2} - 2 \sqrt{3}) + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} } }$

Solution:-

$\frac{4}{(3 \sqrt{2} - 2 \sqrt{3} ) + \frac{3}{3 \sqrt{3 } + 2 \sqrt{2} } } \\ = \frac{4}{3 \times 1.4142 - 2 \times 1.732 + \frac{3}{3 \times 1.732 + 2 \times 4142} } \\ = \frac{4}{4.2426 - 3.464 + \frac{3}{5.196 + 2.8284} } \\ = \frac{4}{0.7786 + \frac{3}{8.0244} } \\ = \frac{4 \times 8.0244} { 6.24779784 + 3 } \\ = \frac{32.0976}{9.24779784} \\ = 3.4708371177 \\ = 3.47(approx)$

Answer 2

Hey Friend...

Here is the Solution..

$\frac{4}{3\sqrt{2}-2\sqrt{3}}+\frac{3}{3\sqrt{3}+2\sqrt{2}} \\\\\frac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{2}-2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{3}+2\sqrt{2})}\\\\\frac{6\sqrt{3}+17\sqrt{2}}{9\sqrt{6}-12-18-4\sqrt{6}}\\ \\\frac{6\sqrt{3}+17\sqrt{2}}{5\sqrt{6}-30}\\\\\frac{6*1.732+17*1.414}{5*\sqrt{2}*\sqrt{3}-30}\\\\\frac{10.392+24.038}{5*1.414*1.732-30}\\\\\frac{34.43}{12.2452-30}\\\\\frac{34.43}{-17.75}\\\\=1.939\\= 2$

Therefore, The answer = 2

.

.

.

.

I Hope This Will Help You