Question

please answer both of them ​

Answer 1

SOLUTION 1 :-

$\star \: \sf \: 2.3 \bar{5} + 0. \bar{12}$

Firstly let's try to convert them in rational form i.e p/q where p and q are integers and q ≠ 0.

Let,

$\implies \: \sf \: x = 2.3 \bar{5} \: and \: y = 0. \bar{12}$

Let's solve both the term seperately,

$\implies \: \sf \: x = 2.3 \bar{5}$

Multiple 10 on both sides,

$\implies \: \sf \: 10x = 23. \bar{5} \: ....(1)$

Again multiply by 10 on both sides,

$\implies \: \sf \: 100x = 235. \bar{5} \: ....(2)$

Now subtract equation 1 from equation 2,

$\implies \: \sf \:100 x - 10x = 235. \bar{5} - 23. \bar{5}$

$\implies \: \sf \: 90x = 212$

$\implies \: \sf \: x = \dfrac{212}{90}$

$\therefore \sf \: \underline{ \boxed{2.3 \bar{5} = \frac{212}{90} }}$

Similarly,

$\implies \: \sf \: y = 0.\bar{12} \: ....(1)$

multiply by 100 on both sides,

$\implies \: \sf \: 100y = 12.\bar{12} \: ....(2)$

subtract equation 1 from equation 2,

$\implies \: \sf \:100 y - y =12. \bar{12} - 0.\bar{12}$

$\implies \: \sf \: 99y = 12$

$\implies \: \sf \: y = \dfrac{12}{99}$

$\therefore \sf \: \underline{ \boxed{0. \bar{12} = \frac{12}{99} }}$

Now According to the question we have,

$\implies \: \sf \: x + y = \dfrac{212}{90} + \dfrac{12}{99}$

$\implies \: \sf \: x + y = \dfrac{2332 +120}{990}$

$\implies \: \sf \: x + y = \dfrac{2452}{990}$

$\implies \: \sf \: x + y = 2.4 \bar{76}$

SOLUTION :- 2

$\implies \: \sf \: \bigg( \sqrt{ \dfrac{3}{5} } \bigg) ^{x + 1} = \dfrac{125}{27}$

By using identify √a¹ = a^½.

$\implies \: \sf \: \bigg( \dfrac{3}{5} \bigg) ^{ \dfrac{x + 1}{2} } = \dfrac{5 ^{3} }{3 ^{3} }$

$\implies \: \sf \: \bigg( \dfrac{3}{5} \bigg) ^{ \dfrac{x + 1}{2} } = \bigg( \dfrac{5}{3} \bigg) ^{3}$

$\implies \: \sf \: \bigg( \dfrac{3}{5} \bigg) ^{ \dfrac{x + 1}{2} } = \bigg( \dfrac{3}{5} \bigg) ^{ - 3}$

$\implies \: \sf \: \dfrac{x + 1}{2} = - 3$

$\implies \: \sf \: x + 1 = - 6$

$\implies \: \sf \: x = - 6 - 1$

$\implies \: \underline{ \boxed{\sf \: x = - 7}}$