Question

$\huge{ \green{ \boxed{ \mathfrak{QUESTIONS!:}}}}$

13- If \: a+\frac{1}{a}=\frac{17}{4}, \: Then\:find \: the \: value \: of \: (a - \frac{1}{a}) \: and \: ( {a}^{2} - \frac{1}{ {a}^{2} })

14- Factorise:
(i) 81(x+1)² + 90(x+1)(y+2) + 25(y+2)²
(ii) x² - 1 - 2y - y²

Pliz solve this guys! :(​

Answer 1

Step-by-step explanation:

Factories form of the expression is 81(x+1)^2+90(x+1)(y+2)+25(y+2)^2=(9x+5y+19)(9x+5y+19)81(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

=(9x+5y+19)(9x+5y+19)

Step-by-step explanation:

Given : Expression 81(x+1)^2+90(x+1)(y+2)+25(y+2)^281(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

To find : Factories the expression?

Solution :

The given expression 81(x+1)^2+90(x+1)(y+2)+25(y+2)^281(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

is in the form of a^2+2ab+b^2a

2

+2ab+b

2

in which

a=9(x+1)a=9(x+1)

b=5(y+2)b=5(y+2)

We know, a^2+2ab+b^2=(a+b)^2a

2

+2ab+b

2

=(a+b)

2

Substitute a and b,

(9(x+1))^2+2(9(x+1))(5(y+2))+(5(y+2))^2=((9(x+1))+(5(y+2)))^2(9(x+1))

2

+2(9(x+1))(5(y+2))+(5(y+2))

2

=((9(x+1))+(5(y+2)))

2

81(x+1)^2+90(x+1)(y+2)+25(y+2)^2=(9x+9+5y+10)^281(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

=(9x+9+5y+10)

2

81(x+1)^2+90(x+1)(y+2)+25(y+2)^2=(9x+5y+19)^281(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

=(9x+5y+19)

2

81(x+1)^2+90(x+1)(y+2)+25(y+2)^2=(9x+5y+19)(9x+5y+19)81(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

=(9x+5y+19)(9x+5y+19)

Therefore, factories form of the expression is 81(x+1)^2+90(x+1)(y+2)+25(y+2)^2=(9x+5y+19)(9x+5y+19)81(x+1)

2

+90(x+1)(y+2)+25(y+2)

2

=(9x+5y+19)(9x+5y+19)

Answer 2

Answer:

Step-by-step explanation: