Question

(2√5+√3)÷(2√5-√3)+(2√5-√3)÷(2√5+√3)=a+√15b
​

Answer 1

Step-by-step explanation:

gi Budd this is your answer

Answer 2

Answer:

The value of a and b is $\frac{46}{17} \ and\ 0$

Step-by-step explanation:

We have

$\frac{2\sqrt{5}+\sqrt{3}}{2\sqrt5-\sqrt{3}}+ \frac{2\sqrt{5}-\sqrt{3}}{2\sqrt5+\sqrt{3}}\\=\frac{2\sqrt{5}+\sqrt{3}}{2\sqrt5-\sqrt{3}} \times \frac{2\sqrt{5}+\sqrt{3}}{2\sqrt5+\sqrt{3}}+\frac{2\sqrt{5}+\sqrt{3}}{2\sqrt5+\sqrt{3}} \times \frac{2\sqrt{5}-\sqrt{3}}{2\sqrt5-\sqrt{3}}$

$=\frac{(2\sqrt{5}+\sqrt{3})^{2}}{(2\sqrt5)^{2}-(\sqrt3)^{2}}}+ \frac{(2\sqrt{5}-\sqrt{3})^{2}}{(2\sqrt5)^{2}-(\sqrt3)^{2}}}\\\\=\frac{(2\sqrt{5})^{2}+2(2\sqrt{5}*\sqrt{3})+(\sqrt{3})^{2}}{(2\sqrt5)^{2}-(\sqrt3)^{2}}}+\frac{(2\sqrt{5})^{2}-2(2\sqrt{5}*\sqrt{3})+(\sqrt{3})^{2}}{(2\sqrt5)^{2}-(\sqrt3)^{2}}}\\\\=\frac{20+4\sqrt{15}+3}{(20-3)}+\frac{20-4\sqrt{15}+3}{(20-3)}}\end{array}$$=\frac{20+4\sqrt{15}+3}{17}+\frac{20-4\sqrt{15}+{3}}{17}}\\\\=\frac{23+4\sqrt{15}-4\sqrt{15}+23}{17}}\\=\frac{46}{17}$

Now $a+\sqrt{15} b\\=\frac{46}{17}+0\\a=\frac{46}{17}\\b=0$