Question

please solve it's urjent
FACTORIES ÷
1) 9-a²+2ab - b²
2) 1+2ab-(a²+b²)
3) x²-y²+6y-9
4) a²-b²+2ab-c²
5) x²-2+ 1/x² - y²
please solve then I will mark brainist and gives 25 pts please solve it ​for class 9

Answer 1

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1)

${\bf{\green{=>9-a^2+2ab-b^2}}}$

${\bf{\green{=>3^2-(a^2-2ab+b^2)}}}$

${\bf{\green{=>3^2-(a-b)^2}}}$

${\bf{\green{=>(3+a-b)(3-a+b)}}}$

2)

${\bf{\pink{=>1+2ab-(a^2+b^2)}}}$

${\bf{\pink{=>1-(a^2+b^2-2ab)}}}$

${\bf{\pink{=>1^2-(a-b)^2}}}$

${\bf{\pink{=>(1+a-b)(1-a+b)}}}$

3)

${\bf{\blue{=>x^2-y^2+6y-9}}}$

${\bf{\blue{=>x^2-(3^2+y^2-6y)}}}$

${\bf{\blue{=>x^2-(y-3)^2}}}$

${\bf{\blue{=>(x+y-3)(x-y+3)}}}$

4)

${\bf{\red{=>a^2-b^2+2ab-c^2}}}$

${\bf{\red{=>(a^2+2ab-b^2)-c^2}}}$

${\bf{\red{=>(a+b)^2-c^2}}}$

${\bf{\red{=>(a+b+c)(a+b-c)}}}$

5)

${\bf{\green{=>x^2-2+1/x^2+y^2}}}$

${\bf{\green{=>x^2-1^2/x^2+y^2}}}$

${\bf{\green{=>(x+1)(x-1)/x^2+y^2}}}$

Answer 2

Step-by-step explanation:

1).

9 - a² + 2ab - b²

=> 3² - ( a² - 2ab + b²)

=> 3² - (a-b)²

=> ( 3- a +b) (3+a-b)

2)

by simplifying

1 +2ab - a² - b² 

1 - (a² - 2ab + b²) 

1 - (a -b)² 

wkt , (a² - b²) = (a+b)(a-b) 

as 1 = 1² 

1² - (a-b)²

= (1+a-b)(1-a+b)

3)

x² - y² + 6y - 9

[Take minus sign common after x²]

= x² - (y² - 6y + 9)

[ Now, we can see a perfect square.

a² - 2ab + b² = (a-b)² .

Here, a = y, b=3]

= x² - (y - 3)²

[ Now, a² - b² = (a+b)(a-b) ]

= (x + (y-3)) (x - (y-3))

= (x + y - 3)(x - y + 3)

4)

Rearrenge the polynominal ;-

a^2+2ab+b^2-c^2

The a and b is a perfect square , with the formula ( a + b)^2 = a^2+2ab+b^2

So, (a+b)^2 - c^2

This leaves you with a difference of squares , whose formula is ( a+b)(a- b)= a^2 - b^2

So, you can give ( a + b+c) ( a-b-c).

5)

x^2 - 2 + 1/x^2 - y^2

= x^2 - 1/(x + y) (x - y)

= x^2 - 1^2/(x + y) (x - y)

= (x + 1) (x - 1)/(x + y) (x - y)

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❣️THANKS KARDO YAR❣️