Question

Question 1
(a) Find the product: (3a + 8) (3a - 8)
(b) Prove that
(x-1+y)(x + y)-1= 1/xy

Question 2
(a) Solve the following equation for n:
2n+1 = 41n-8
(b) Find the value of a² + b² when a + b = 8, ab = 20
Question 3
(a) By using suitable identity, evaluate: 101 x 99


​

Answer 1

Answer:

BRO I ONLY KNOW Q1. & O3.

MARK ME AS BRAINLIEST

Q1.

a. 9a^2+6a-24a-16

    9a^2-18a-16

b. 1/y-1/x∝1/x-y

Or (x-y)/xy=k×1/(x-y)(where k≠0)

Or (x-y)²=kxy

Or x²-2xy+y²=kxy

Or x²+y²=(k+2)xy

Or (x²+y²)/xy=k+2

Or x/y+y/x=k+2

Or (x/y)²+1=c(x/y) (where c=k+2 =constant)

Or (x/y)²-2(x/y)(c/2)+c²/4+1-c/4=0

Or (x/y-c/2)²=(c²/4)–1

Or x/y-c/2=±√(c²-4)/2

Or x/y=(c/2)±√(c²-4)/2

Or x/y=constant

Therefore x∝y

Q3.

101×99.

=(100+1)(100-1).

by using the property (a+b)(a-b)=a²-b²

=(100)²-1

=10000-1

=9999

Answer 2

answer1:

(a)3a sq-8 sq [(a+b) (a-b) =a sq-b sq]

(b)don't know

answer2:

(a)2n+1=41n-8

=8+1=41n-2n

=9=39n

=n=39/9

(b)question is wrong it must be 9 at the place 8.

answer3:

(a)(100+1)99

=9900+99

=9999