Question

Y'ALL HELP PLEASE
Simplify by rationalising the denominator​

Answer 1

Ans:- (√7-√5)/(√7+√5)

= (√7-√5)/(√7+√5)×(√7-√5)/(√7-√5)

= (7 + 5 - 2√35)/7-5

= (12 - 2√35)/2

= 2(6-√35)/2

= 6 - √35

2. 1/(3√2-2√3)

= 1/(3√2-2√3)×(3√2+2√3)/(3√2+2√3)

= (3√2 + 2√3)/(18-12)

= 3√2/6 + 2√3/6

3. (3√5 - √7)/(3√3 + √2)

= (3√5 - √7)/(3√3 + √2) × (3√3 - √2)/(3√3 - √2)

= (9√15 - 3√10 + √14 - 3√21)/(27 - 2)

= 9√15/25 - 3√10/25 + √14/25 - 3√21/25

here is your answer

hope it will help you

Answer 2

Answer:

(i) 6 - √35

(ii) [3√2 + 2√3]/6

(iii) [9√15 - 3√10 - 3√21 + √14]/25

Step-by-step explanation:

(i)

$\frac{\sqrt{7}-\sqrt{5} }{\sqrt{7}+\sqrt{5} } \\\\\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}} *\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}} \\\\\frac{(\sqrt{7}-\sqrt{5})^{2} }{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})} \\\\\frac{(\sqrt{7} )^{2}+(\sqrt{5})^{2} -2(\sqrt{7} )(\sqrt{5} )}{(\sqrt{7} )^{2} -(\sqrt{5} )^{2}} \\\\\frac{7+5-2\sqrt{35} }{7-5} \\\\\frac{12-2\sqrt{35} }{2} \\\\\frac{2(6-\sqrt{35} )}{2} \\\\6-\sqrt{35}$

(ii)

$\frac{1}{3\sqrt{2} -2\sqrt{3} } \\\\\frac{1}{3\sqrt{2} -2\sqrt{3}} *\frac{3\sqrt{2} +2\sqrt{3}}{3\sqrt{2} +2\sqrt{3}} \\\\\frac{3\sqrt{2} +2\sqrt{3}}{(3\sqrt{2} )^{2}-(2\sqrt{3} )^{2} } \\\\\frac{3\sqrt{2} +2\sqrt{3}}{9(2)-4(3)} \\\\\frac{3\sqrt{2} +2\sqrt{3}}{18-12} \\\\\frac{3\sqrt{2} +2\sqrt{3}}{6}$

(iii)

$\frac{3\sqrt{5}-\sqrt{7} }{3\sqrt{3} +\sqrt{2} }\\\\\frac{3\sqrt{5}-\sqrt{7} }{3\sqrt{3} +\sqrt{2} }*\frac{3\sqrt{3} -\sqrt{2} }{3\sqrt{3} -\sqrt{2} } \\\\\frac{(3\sqrt{5}-\sqrt{7})(3\sqrt{3}-\sqrt{2} ) }{(3\sqrt{3})^{2} -(\sqrt{2} )^{2} } \\\\\frac{9\sqrt{15}-3\sqrt{10}-3\sqrt{21} +\sqrt{14} }{27-2} \\\\\frac{9\sqrt{15}-3\sqrt{10}-3\sqrt{21} +\sqrt{14}}{25}$