Question

please please please solve it step by step ​

Answer 1

Answer:

Hope it helps you.........

Answer 2

$\red\bigstar$ Answer:

• $\sf \boxed{ \sf \: x \: - \: \dfrac{1}{x} \: = \: \pm 10 }$

$\pink\bigstar$ Given:

• $\sf \: {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 102$

$\blue\bigstar$ To find:

• $\sf \: x \: - \: \dfrac{1}{x}$

$\red\bigstar$ Solution:

We know that :

• (x - y)² = x² - 2xy + y²

Now,

x² + y² can be written as (x - y)² + 2xy

Similarly,

$\sf \: {x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: {( x \: - \: { \dfrac{1}{x} })}^{2} \: + \: 2 \times x \times \dfrac{1}{x}$

By substituting the values, we get:

$\sf\implies \: 102 \: = \: {( x \: - \: { \dfrac{1}{x} })}^{2} \: + \: 2$

$[ \sf\because x \: and \: \dfrac{1}{x} \: get \: cancelled]$

$\sf\implies \: 102 \: - \: 2 \: = {( x \: - \: { \dfrac{1}{x} })}^{2}$

[ By transposition method ]

$\sf\implies \: 100 \: = \: {( x \: - \: { \dfrac{1}{x} })}^{2}$

$\sf\implies \: \sqrt{100} \: = \: x \: - \: \dfrac{1}{x}$

$\sf\implies \boxed{ \sf\: \pm 10 \: = \: x \: - \: \dfrac{1}{x} }$

$\sf\boxed{ \sf\therefore\: \: x \: - \: \dfrac{1}{x} = \: \pm 10 }$

$\pink\bigstar$ Concepts Used:

• (x - y)² = x² - 2xy + y²
• x² + y² = (x - y)² + 2xy
• Substitution of values
• Cancellation of the same number multipled and divided again
• Transposition Method

$\green\bigstar$Extra - Information:

• (a + b)² = a² + 2ab + b²

• (a – b)² = a² – 2ab + b²

• a² – b² = (a + b)(a – b)

• (y + a)(y + b) = y² + (a + b)y + ab

• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

• (a + b)³ = a³ + b³ + 3ab (a + b)

• (a – b)³ = a³ – b³ – 3ab (a – b)

• a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

• If (a + b + c ) = 0, then a³ + b³ + c³ = 3abc