Question

pls answer this question

Answer 1

Answer:

1

Step-by-step explanation:

Given:

a + 1/b = 3    ---- (1)

b + 1/c = 4    ----- (2)

c + 1/a = 9/11    ----- (3)

Equation (1) can be written as,

⇒ a + 1/b = 3

⇒ a = 3 - (1/b)

⇒ a = (3b - 1)/b

Then, 1/a = (b)/(3b - 1)   ---- (4)

Equation (2) Can be written as,

⇒ b + 1/c = 4

⇒ 1/c = 4 - b

⇒ c = 1/(4 - b)    ----- (5)

Equation (4) & (5) can be solved as,

⇒ (1/a) + c = (b/3b - 1) + (1/4 - b)

⇒ 9/11 = (4b - b² + 3b - 1)/(12b - 4 - 3b² + b)

⇒ 9/11 = (7b - b² - 1)/(13b - 3b² - 4)

⇒ 9(13b - 3b² - 4) = 11(7b - b² - 1)

⇒ 117b - 27b² - 36 = 77b - 11b² - 11

⇒ -16b² + 40b - 25 = 0

⇒ 16b² - 40b + 25 = 0

⇒ (4b - 5)² = 0

⇒ 4b = 5

⇒ b = 5/4

Then, Substitute b = 5/4 in (1), we get

⇒ a + 1/b = 3

⇒ a = 3 - 1/b

⇒ a = 3 - 4/5

⇒ a = 11/5.

Substitute b = 5/4 in (5), we get

⇒ c = 1/4 - b

⇒ c = 1/(4 - 5/4)

⇒ c = 4/11.

Now,

abc = (5/4) * (11/5) * (4/11)

      = 1.

Hope it helps!

Answer 2

Step-by-step explanation:

a+1/b=3 , b+1/c=4 , c+1/a =9/11

(a+1/b)(b+1/c)=3×4

or, (ab+a/c+1+1/bc)=12

or, ab+a/c+1/bc=12–1=11

(c+1/a)(ab+a/c+1/bc)=(9/11)×11

or, abc+a+1/b+b+1/c+1/abc=9

or, abc+3+4+1/abc=9

or, abc+(1/abc)=9–7=2

or, {( abc)^2+1}/abc=2

or, ( abc)^2+1=2abc

or, (abc)^2-2abc+1=0

or, ( abc-1)^2=0

or, abc-1=0

or, abc=1