Question

pls solve the circled ones

Answer 1

7 solution
7+3√5/7-3√5=a+b√5
LHS
we need to rationalize tge denominator
7+3√5/7-3√5
rationalising factor is 7+3√5
7+3√5÷7-3√5×7+3√5÷7+3√5
(7+3√5)^2÷(7+3√5)(7-3√5)
(a+b)(a-b)=a^2-b^2
(a+b)^2=a^2+2ab+b^2
49+42√5+45÷49-45
94+42√5÷4
(94÷4)+(42√5)÷4
(47/2)+(21)√5=a+b√5
a=47÷2
b=21