Question

three question from 9 th std cbse. plz ans.​

Answer 1

Step-by-step explanation:

11. (x+1)³-(x-1)³

(a+b)³ =a³+b³+3ab(a+b)

(a-b)³=a³-b³-3ab(a-b)

=(x³+(1)³+3x(x+1)-x³-(1)³-3x(x-1)

=x³+1+3x²+3x-x³-1-3x²+3x

=6x

12. √5-2/√5+2-√5+2/√5-2

=√5-2/√5+2×√5-2/√5-2-√5+2/√5-2×√5+2/√5+2

={(√5-2)²/(√5+2)(√5-2)}-{(√5+2)²/(√5-2)×(√5+2)}

={(√5)²+(2)²-2×√5×2/(√5)²-2√5+2√5-4}-{(√5)²+(2)²+2√5×2/(√5)²+2√5-2√5-4}

={5+4-4√5/5-4}-{5+4+4√5/5-4}

=9-4√5-9+4√5

=0

13. (x-1)=0. and. (x+3)=0

=x=1. x=-3

Now,

x³-ax²-13x+b=0

put x=1

(1)³-a(1)²-13×1+b=0

1-a-13+b=0

-a+b=12

b=12+a. (1)

again,

put x =-3

(-3)³-a(-3)²-13×(-3)+b=0

-27-3a+39+b=0

-3a+b+12=0

-3a+12+a+12=0. (using1)

-2a=-24

a=12

put in (1)

b=12+12

=24