Question

help me
$3 - 2 \sqrt{2} \ - \frac{1}{3 - 2 \sqrt{2} }$
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Answer 1

Given:-

• $3 - 2 \sqrt{2} - \dfrac{1}{3 - 2 \sqrt{2} }$

We have to find the value of this.

Answer:-

Given that,

$3 - 2 \sqrt{2} - \dfrac{1}{3 - 2 \sqrt{2} }$

First we have to simplify the value of $\dfrac{1}{3 - 2\sqrt{2}}$ by rationalising the denominator.

So,

$\dfrac{1}{3 - 2 \sqrt{2} }$

The rationalising factor will be the conjugate pair of the denominator.

So, here Rationalising factor = $3 + 2\sqrt{2}$

Multiplying both numerator and denominator with $3 + 2\sqrt{2},$

${ \longrightarrow \dfrac{1}{3 - 2 \sqrt{2}} = \dfrac{1(3 + 2\sqrt{2}) }{(3 - 2 \sqrt{2})(3 + 2 \sqrt{2}) } }$

Applying (a - b)(a + b) = a² - b² with a = 3 and b = 2√2,

$\longrightarrow \dfrac{1}{3 - 2 \sqrt{2} } = \dfrac{3 + 2\sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} }$

$\longrightarrow \dfrac{1}{3 - 2 \sqrt{2} } = \dfrac{3 + 2\sqrt{2} }{ 9 - 8 }$

$\longrightarrow \dfrac{1}{3 - 2 \sqrt{2} } = \dfrac{3 + 2\sqrt{2} }{ 1 }$

${ \longrightarrow \dfrac{1}{3 - 2 \sqrt{2} } = 3 + 2\sqrt{2} \quad \dots(1)}$

So,

$3 - 2 \sqrt{2} - \dfrac{1}{3 - 2 \sqrt{2} }$

From (1),

$= 3 - 2 \sqrt{2} - (3 + 2\sqrt{2})$

$= 3 - 2 \sqrt{2} - 3 - 2\sqrt{2}$

$= - 4\sqrt{2}$

Hence,

${ \longrightarrow \underline{ \underline{3 - 2 \sqrt{2} - \dfrac{1}{3 - 2 \sqrt{2} } = - 4 \sqrt{2} }}}$