Question

rationalize a denominator of the following: √3-1\√3+1.
please tell me​

Answer 1

Answer:

▶2 + √3

Explanation:

Given that,

▶ √3 - 1/√3 + 1

▶ √3 - 1/√3 + 1 × √3 - 1 /√3 - 1

▶ (√3 - 1)(√3-1)/(√3 + 1)(√3 - 1)

▶ (√3 + 1)²/(√3)² - (1)²

▶ 3 + 2√3 + 1/3 - 1

▶ 4 + 2√3 /2

▶ 2(2 + √3)/2

▶ 2 + √3

Hence,

• √3-1/√3+1 = 2+√3.

Some Algebraic Identities :

• (a + b)² = a² + b² + 2ab²

• (a - b)² = a² - b² + 2ab

• (a² - b²) = (a + b) (a - b)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ + 3ab(a + b)

• (x + a) (x + b) = x² + (a+b)x + ab

Answer 2

$\frac{√3 \: - 1}{√3 + 1}$

$= \frac{√3 \: - 1 }{√3 + 1 } \frac{ \times }{ \times } \frac{√3 \: - 1}{√3 - 1}$

$= \frac{(√3 - 1)^{2} }{(√3)^{2} - (1)^{2} }$

$= \frac{3 \: \: + \: \: 1 \: \: \: - 2 \: \: \: √3}{3 - 1}$

$= \frac{4 - 2 \: \: √3}{2}$

$= \frac{2(2 - √3)}{2}$

Numerator - 2 and denominator 2 are canceled

$\huge\red{\therefore} 2 - √3$