Using the formula (a+b)(a b) = a2 b2 (a + x) (a x) (a2 x2) = [(a)2 (x)2 ] (a2 x2) = (a2 x2 ) (a2 x2) = (a2 x2)2 [ as we know (a b)( ...

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Factors Affecting Friction

a+xa-xa2-x2

Question

(a + x) (a - x) (a² - x²)

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Solution

Using the formula (a+b)(a-b) = a² - b²

(a + x) (a - x) (a² - x²)

= [(a)²- (x)²] (a² - x²)

= (a² - x²) (a² - x²)

= (a² - x²)² [ as we know (a-b)(a-b) = (a-b)² ]

= (a² )²- 2a²x²+ (x²)² [ as (a-b)²= (a)² -2a²b²+ (b)² ]

= a⁴- 2a²x²+ x⁴Ans.

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\frac{x}{a} + \frac{y}{b} = (a + b), \frac{x}{a^{2}} + \frac{y}{b^{2}} = 2 .

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} {(1 + x)}^{a} & {(2 + x)}^{b} & 1 1 & {(1 + x)}^{a} & {(2 + x)}^{b} {(2 + x)}^{b} & 1 & {(1 + x)}^{a} \end{matrix} ∣ ∣ ∣ ∣ ∣

f (x)

\tan^{- 1} \{\frac{x}{\sqrt{a^{2} - x^{2}}}\}, - a < x < a

\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1

{[\{\frac{x^{a (a - b)}}{x^{a (a + b)}}\} \div \{\frac{x^{b (b - a)}}{x^{b (b + a)}}\}]}^{a + b} = 1

{(x^{\frac{1}{a - b}})}^{\frac{1}{a - c}} {(x^{\frac{1}{b - c}})}^{\frac{1}{b - a}} {(x^{\frac{1}{c - a}})}^{\frac{1}{c - b}} = 1

{(\frac{x^{a^{2} + b^{2}}}{x^{a b}})}^{a + b} {(\frac{x^{b^{2} + c^{2}}}{x^{b c}})}^{b + c} {(\frac{x^{c^{2} + a^{2}}}{x^{a c}})}^{a + c} = x^{2 (a^{3} + b^{3} + c^{3})}

{(x^{a - b})}^{a + b} {(x^{b - c})}^{b + c} {(x^{c - a})}^{c + a} = 1

{\{{(x^{a - a^{- 1}})}^{\frac{1}{a - 1}}\}}^{\frac{a}{a + 1}} = x

{(\frac{a^{x + 1}}{a^{y + 1}})}^{x + y} {(\frac{a^{y + 2}}{a^{z + 2}})}^{y + z} {(\frac{a^{z + 3}}{a^{x + 3}})}^{z + x} = 1

{(\frac{3^{a}}{3^{b}})}^{a + b} {(\frac{3^{b}}{3^{c}})}^{b + c} {(\frac{3^{c}}{3^{a}})}^{c + a} = 1

Factorise :

$x^{2} - x (a + 2 b) + 2 a b$

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