Simplify the following: Class 9

Simplify the following: Class 9

Question 1.

Simplify the following: Class 9

i. $\quad \frac{4 x^2+4 x+1}{4 x^2-1}$

ii. $\quad \frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}$

iii. $\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}$

Note: Assume that the denominators are not equal to 0.

Solution:

i. $\begin{aligned} & \frac{4 x^2+4 x+1}{4 x^2-1}=\frac{(2 x)^2+2(2 x)(1)+1^2}{(2 x)^2-1^2} \\ & =\frac{(2 x+1)^2}{(2 x+1)(2 x-1)} \\ & \quad \ldots\left[\because(a+b)^2=a^2+2 a b+b^2,\right. \\ & \left.\quad a^2-b^2=(a+b)(a-b)\right] \\ & =\frac{(2 x+1)(2 x+1)}{(2 x+1)(2 x-1)}=\frac{2 x+1}{2 x-1}\end{aligned}$

......[Cancelling the common factor (2x + 1)]

ii. $\begin{aligned} & \frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}=\frac{9 \times 3\left(a^3-8 b^3\right)}{9\left(a^2-4 b^2\right)} \\ & =\frac{3\left[a^3-(2 b)^3\right]}{a^2-(2 b)^2} \\ & =\frac{3(a-2 b)\left(a^2+(a)(2 b)+(2 b)^2\right)}{(a+2 b)(a-2 b)}\end{aligned}$

...[∵ A³ - B³ = (A - B)( A² + AB + B²), A² - B² = (A + B)(A - B)]

$=\frac{3\left(a^2+2 a b+4 b^2\right)}{a+2 b}$

...[Cancelling the common factor (a - 2b)]

iii. $\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}$

s³ + 125t³ = s³ + (5t)³

= (s + 5t)(s² - (s)(5t) + (5t)²)

.....[∵ a³ + b³ = (a + b)(a² - ab + b²)]

= (s + 5t)(s² - 5st + 25t²)

s² - 2st - 35t²

Since 5 + (-7) = -2 and 5 × (-7) = -35, the numbers are 5 and -7.

∴ s² - 2st - 35t² = (s + 5t)(s - 7t)

$\begin{aligned} \therefore \quad & \frac{s^3+125 t^3}{s^2-2 s t-35 t^2} \\ & =\frac{(s+5 t)\left(s^2-5 s t+25 t^2\right)}{(s+5 t)(s-7 t)} \\ & =\frac{s^2-5 s t+25 t^2}{s-7 t}\end{aligned}$

... [Cancelling the common factor (s + 5t)]