Question

show that 1/(2-root3)-1/(root 3-root 2)+1/(root 2-1)=3​

Answer 1

Step-by-step explanation:

As per the provided information in the given question, we have to prove that,

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

In order to prove this, firstly we need to find the exact value after performing simplification in L.H.S.

Solving L.H.S :

We've to rationalise the denominator of the terms in L.H.S to performing addition and subtraction.

$\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} \quad \dots (Expression \; 1) }$

In order to rationalize the denominator, we multiply the rationalising factor of the denominator of the fraction with both the numerator and the denominator of the fraction. Rationalising factor of (2 - √3) is (2 + √3).

Expression 1 :

Multiplying (2 + √3) with both the numerator and the denominator of the fraction.

$\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2+\sqrt{3}} }$

Rearranging the terms.

$\longmapsto \rm { \dfrac{1(2 + \sqrt{3})}{(2-\sqrt{3})(2 + \sqrt{3})} }$

Using (a + b)(a - b) = a² - b² Identity.

$\longmapsto \rm { \dfrac{2 + \sqrt{3}}{(2)^2-(\sqrt{3})^2} }$

Writing the squares of the numbers in the denominator.

$\longmapsto \rm { \dfrac{2 + \sqrt{3}}{4-3} }$

Performing subtraction in the denominator.

$\longmapsto \rm { \dfrac{2 + \sqrt{3}}{1} }$

Now, this expression can be written as,

$\longmapsto \bf { 2 + \sqrt{3} }$

Expression 2 :

Rationalising factor of (√3 - √2) is (√3 + √2). Multiplying (√3 + √2) with both the numerator and the denominator of the fraction.

$\longmapsto \rm { \dfrac{1}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3}+\sqrt{2}} }$

Rearranging the terms.

$\longmapsto \rm { \dfrac{1(\sqrt{3} + \sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3} + \sqrt{2})} }$

Using (a + b)(a - b) = a² - b² Identity.

$\longmapsto \rm { \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} }$

Writing the squares of the numbers in the denominator.

$\longmapsto \rm { \dfrac{\sqrt{3} + \sqrt{2}}{3-2} }$

Performing subtraction in the denominator.

$\longmapsto \rm { \dfrac{\sqrt{3} + \sqrt{2}}{1} }$

Now, this expression can be written as,

$\longmapsto \bf { \sqrt{3} + \sqrt{2} }$

Expression 3 :

Rationalising factor of (√2 - √1) is (√2 + √1). Multiplying (√2 + √1) with both the numerator and the denominator of the fraction.

$\longmapsto \rm { \dfrac{1}{\sqrt{2}-1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2}+1} }$

Rearranging the terms.

$\longmapsto \rm { \dfrac{1(\sqrt{2} +1)}{(\sqrt{2}-1)(\sqrt{2} +1)} }$

Using (a + b)(a - b) = a² - b² Identity.

$\longmapsto \rm { \dfrac{\sqrt{2} + 1}{(\sqrt{2})^2-(1)^2} }$

Writing the squares of the numbers in the denominator.

$\longmapsto \rm { \dfrac{\sqrt{2} + 1}{2-1} }$

Performing subtraction in the denominator.

$\longmapsto \rm { \dfrac{\sqrt{2} + 1}{1} }$

Now, this expression can be written as,

$\longmapsto \bf { \sqrt{2} + 1 }$

Adding all the terms of the L.H.S :

$\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} - \dfrac{1}{\sqrt{3}-\sqrt{2}} + \dfrac{1}{\sqrt{2} - 1}}$

Substitute the simplified values.

$\longmapsto \rm { (2 + \sqrt{3}) - (\sqrt{3} + \sqrt{2} ) + ( \sqrt{2} +1 ) }$

Removing the brackets.

$\longmapsto \rm { 2\cancel{ + \sqrt{3} - \sqrt{3} } \cancel{- \sqrt{2} ) + \sqrt{2} } + 1 }$

Performing addition and subtraction.

$\longmapsto \rm { 2 + 1}$

$\longmapsto \bf { 3 \quad\dots L.H.S.}$

In the R.H.S, we have :

$\longmapsto \bf { 3 \quad\dots R.H.S.}$

On comparing L.H.S and R.H.S, we get that,

• L.H.S (3) = R.H.S (3)

Hence, proved!

Answer 2

Answer:

As per the provided information in the given question, we have to prove that,

\begin{gathered}\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} - \dfrac{1}{\sqrt{3}-\sqrt{2}} + \dfrac{1}{\sqrt{2} - 1} = 3} \\\end{gathered}

⟼

2−

3

1

−

3

−

2

1

+

2

−1

1

=3

In order to prove this, firstly we need to find the exact value after performing simplification in L.H.S.

Solving L.H.S :

We've to rationalise the denominator of the terms in L.H.S to performing addition and subtraction.

\begin{gathered}\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} \quad \dots (Expression \; 1) } \\\end{gathered}

⟼

2−

3

1

…(Expression1)

In order to rationalize the denominator, we multiply the rationalising factor of the denominator of the fraction with both the numerator and the denominator of the fraction. Rationalising factor of (2 - √3) is (2 + √3).

Expression 1 :

Multiplying (2 + √3) with both the numerator and the denominator of the fraction.

\begin{gathered}\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2+\sqrt{3}} } \\\end{gathered}

⟼

2−

3

1

×

2+

3

2+

3

Rearranging the terms.

\begin{gathered}\longmapsto \rm { \dfrac{1(2 + \sqrt{3})}{(2-\sqrt{3})(2 + \sqrt{3})} } \\\end{gathered}

⟼

(2−

3

)(2+

3

)

1(2+

3

)

Using (a + b)(a - b) = a² - b² Identity.

\begin{gathered}\longmapsto \rm { \dfrac{2 + \sqrt{3}}{(2)^2-(\sqrt{3})^2} } \\\end{gathered}

⟼

(2)

2

−(

3

)

2

2+

3

Writing the squares of the numbers in the denominator.

\begin{gathered}\longmapsto \rm { \dfrac{2 + \sqrt{3}}{4-3} } \\\end{gathered}

⟼

4−3

2+

3

Performing subtraction in the denominator.

\begin{gathered}\longmapsto \rm { \dfrac{2 + \sqrt{3}}{1} } \\\end{gathered}

⟼

1

2+

3

Now, this expression can be written as,

\begin{gathered}\longmapsto \bf { 2 + \sqrt{3} } \\\end{gathered}

⟼2+

3

Expression 2 :

Rationalising factor of (√3 - √2) is (√3 + √2). Multiplying (√3 + √2) with both the numerator and the denominator of the fraction.

\begin{gathered}\longmapsto \rm { \dfrac{1}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3}+\sqrt{2}} } \\\end{gathered}

⟼

3

−

2

1

×

3

+

2

3

+

2

Rearranging the terms.

\begin{gathered}\longmapsto \rm { \dfrac{1(\sqrt{3} + \sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3} + \sqrt{2})} } \\\end{gathered}

⟼

Using (a + b)(a - b) = a² - b² Identity.

\begin{gathered}\longmapsto \rm { \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} } \\\end{gathered}

Writing the squares of the numbers in the denominator.

\begin{gathered}\longmapsto \rm { \dfrac{\sqrt{3} + \sqrt{2}}{3-2} } \\\end{gathered}

Performing subtraction in the denominator.

\begin{gathered}\longmapsto \rm { \dfrac{\sqrt{3} + \sqrt{2}}{1} } \\\end{ggathered}

Now, this expression can be written as,

\begin{gathered}\longmapsto \bf { \sqrt{3} + \sqrt{2} } \\\end{gathered}

Expression 3 :

Rationalising factor of (√2 - √1) is (√2 + √1). Multiplying (√2 + √1) with both the numerator and the denominator of the fraction.

\begin{gathered}\longmapsto \rm { \dfrac{1}{\sqrt{2}-1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2}+1} } \\\end{gathered}

Rearranging the terms.

\begin{gathered}\longmapsto \rm { \dfrac{1(\sqrt{2} +1)}{(\sqrt{2}-1)(\sqrt{2} +1)} } \\\end{ggathered}

Using (a + b)(a - b) = a² - b² Identity.

\begin{gathered}\longmapsto \rm { \dfrac{\sqrt{2} + 1}{(\sqrt{2})^2-(1)^2} } \\\end{gathered}

Writing the squares of the numbers in the denominator.

\begin{gathered}\longmapsto \rm { \dfrac{\sqrt{2} + 1}{2-1} } \\\end{gathered}

Performing subtraction in the denominator.

\begin{gathered}\longmapsto \rm { \dfrac{\sqrt{2} + 1}{1} } \\\end{gathered}

Now, this expression can be written as,

\begin{gathered}\longmapsto \bf { \sqrt{2} + 1 } \\\end{gathered}

Adding all the terms of the L.H.S :

\begin{gathered}\longmapsto \rm { \dfrac{1}{2-\sqrt{3}} - \dfrac{1}{\sqrt{3}-\sqrt{2}} + \dfrac{1}{\sqrt{2} - 1}} \\\end{gathered}

Substitute the simplified values.

\begin{gathered}\longmapsto \rm { (2 + \sqrt{3}) - (\sqrt{3} + \sqrt{2} ) + ( \sqrt{2} +1 ) } \\\end{gathered}

Removing the brackets.

\begin{gathered}\longmapsto \rm { 2\cancel{ + \sqrt{3} - \sqrt{3} } \cancel{- \sqrt{2} ) + \sqrt{2} } + 1 } \\\end{gathered}

Performing addition and subtraction.

\begin{gathered}\longmapsto \rm { 2 + 1} \\\end{gathered}

⟼2+1

\begin{gathered}\longmapsto \bf { 3 \quad\dots L.H.S.} \\\end{gathered}

⟼3…L.H.S.

In the R.H.S, we have :

\begin{gathered}\longmapsto \bf { 3 \quad\dots R.H.S.} \\\end{gathered}

⟼3…R.H.S.

On comparing L.H.S and R.H.S, we get that,

L.H.S (3) = R.H.S (3)

Hence, proved!