Question

please give answer fast........

Answer 1

Given,

 x = 1 / ( 2 - √3 )

To rationalize the denominator we shall multiply numerator and denominator of R.H.S by ( 2 + √3 ).

⇒ x = 1 ( 2 + √3 ) ÷ ( 2 - √3 ) ( 2 + √3 )

⇒ x = ( 2 + √3 ) ÷ { 2² - ( √3 )² }

⇒ x = ( 2 + √3 ) ÷ ( 4 - 3 )

 ∴  x = ( 2 + √3 ).

Now,

⇒ x³ = ( 2 + √3 )³

⇒ x³ = 2³ + ( √3 )³ + 3 × ( 2 ) × ( √3 )² + 3 × ( 2 )² × ( √3 )

⇒ x³ = 8 + 3√3 + 18 + 12√3

⇒ x³ = 26 + 15√3.

Now,

⇒ x² = ( 2 + √3 )²

⇒ x² = 2² + ( √3 )² + 2 × 2 × √3

⇒ x² = 4 + 3 + 4√3

⇒ x² = 7 + 4√3


Now,

= x³ - 2 x² - 7 x + 5

By substituting the values of x³ , x² and x.

= ( 26 + 15√3 ) - 2 ( 7 + 4√3 ) - 7 ( 2 + √3 ) + 5

= 26 + 15√3 - 14 - 8√3 - 14 - 7√3 + 5

= 26 - 14 - 14 + 5 + 15√3 - 8√3 - 7√3

= 31 - 28 + 15√3 - 15√3

= 3.

The final answer is 3.

                          ▬▬▬▬▬▬▬▬▬▬▬

Given ,

 √2 = 1.414 and √6 = 2.449

Now,

= 1 ÷ ( √3 - √2 - 1 )

= 1 ÷ { ( √3 - √2 ) - 1 }

To rationalize the denominator we shall multiply the numerator and denominator by { ( √3 - √2 ) + 1 }.

= 1 { ( √3 - √2 ) + 1 } ÷ { ( √3 - √2 ) - 1 } { ( √3 - √2 ) + 1 }

= ( √3 - √2 + 1 ) ÷ { ( √3 - √2 )² - 1² }

= ( √3 - √2 + 1 ) ÷ { (√3)² + ( √2)² - 2 × √3 × √2 - 1 }

= ( √3 - √2 + 1 ) ÷ ( 3 + 2 - 1 - 2√6 )

= ( √3 - √2 + 1 ) ÷ ( 5 - 1 - 2√6 )

= ( √3 - √2 + 1 ) ÷ ( 4 - 2√6 )

To rationalize the denominator we shall multiply numerator and denominator by ( 4 + 2√6 ).

= ( √3 - √2 + 1 ) ( 4 + 2√6 ) ÷ ( 4 - 2√6 ) ( 4 + 2√6 )

= ( 4√3 + 2√18 - 4√2 - 2√12 + 4 + 2√6 ) ÷ { (4)² - (2√6)² }

= ( 4√3 + 6√2 - 4√2 - 4√3 + 4 + 2√6 ) ÷ ( 16 - 24 )

= ( 2√2 + 2√6 + 4 ) ÷ ( -8 )

= 2 ( √2 + √6 + 2 ) ÷ ( - 8 )

= ( √2 + √6 + 2 ) ÷ ( - 4 )

= - ( √2 + √6 + 2 )÷ 4

By putting the value of √2 and √6.

= - ( 1.414 + 2.449 + 2 ) ÷ 4

= - ( 5.863 ) ÷ 4

= -1.465.

The final answer is ( -1.465 ).