Question :- 15
Simplify :-
[tex]\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} } } - \dfrac{2}{ \sqrt{ {a}^{3} } } - \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
Consider
[tex]\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} }} \\ [/tex]
[tex]\rm \: = \:\dfrac{a - \dfrac{1}{ {a}^{2} } }{ \sqrt{a} - \dfrac{1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \:\dfrac{\dfrac{ {a}^{3} - 1}{ {a}^{2} } }{ \dfrac{a - 1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{3} - 1}{ {a}^{2} } \times \dfrac{ \sqrt{a} }{a - 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{3} - {1}^{3} }{a \times a} \times \dfrac{ \sqrt{a} }{a - 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ \cancel{(a - 1)}( {a}^{2} + 1 + a)}{a \times \sqrt{a} \times \cancel{\sqrt{a} }} \times \dfrac{ \cancel{\sqrt{a}} }{\cancel{a - 1}} \\ [/tex]
[tex]\rm \: = \: \dfrac{{a}^{2} + 1 + a}{ {( \sqrt{a} )}^{2} \times \sqrt{a}} \\ [/tex]
[tex]\rm \: = \: \dfrac{{a}^{2} + 1 + a}{ {( \sqrt{a} )}^{3}} \\ [/tex]
Now, Consider
[tex]\rm \: \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } \\ [/tex]
[tex]\rm \: = \:\dfrac{1 - \dfrac{1}{ {a}^{2} } }{ \sqrt{a} + \dfrac{1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \:\dfrac{\dfrac{ {a}^{2} - 1}{ {a}^{2} } }{ \dfrac{a + 1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{2} - 1}{ {a}^{2} } \times \dfrac{ \sqrt{a} }{a + 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{2} - {1}^{2} }{a \times a} \times \dfrac{ \sqrt{a} }{a + 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ \cancel{(a + 1)}(a - 1)}{a \times \sqrt{a} \times \cancel{\sqrt{a} }} \times \dfrac{ \cancel{\sqrt{a}} }{\cancel{a + 1}} \\ [/tex]
[tex]\rm \: = \: \dfrac{a- 1}{ {( \sqrt{a} )}^{2} \times \sqrt{a}} \\ [/tex]
[tex]\rm \: = \: \dfrac{a- 1}{ {( \sqrt{a} )}^{3}} \\ [/tex]
Now, Consider the given expression, we have
can be rewritten as
[tex]\rm \: = \: \dfrac{{a}^{2} + 1 + a}{ {( \sqrt{a} )}^{3}} - \dfrac{2}{{( \sqrt{a} )}^{3}} - \dfrac{a - 1}{{( \sqrt{a} )}^{3}} \\ [/tex]
[tex]\rm \: = \: \dfrac{1}{{( \sqrt{a} )}^{3}} \bigg( {a}^{2} + a + 1 - 2 - a + 1 \bigg) [/tex]
[tex]\rm \: = \: \dfrac{1}{{( \sqrt{a} )}^{3}} \bigg( {a}^{2} + \cancel{a} + \cancel{2} - \cancel{2} - \cancel{a} \bigg) [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{2} }{{( \sqrt{a} )}^{3}} [/tex]
[tex]\rm \: = \: \dfrac{{( \sqrt{a} )}^{4}}{{( \sqrt{a} )}^{3}} \\ [/tex]
[tex]\rm \: = \: \sqrt{a} \\ [/tex]
Hence,
[tex]\rm\implies \:\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} } } - \dfrac{2}{ \sqrt{ {a}^{3} } } - \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } = \sqrt{a} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Identities Used :-
[tex]\boxed{\rm{ \:x = \sqrt{x} \times \sqrt{x} \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} +xy + {y}^{2}) \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{m} \times {x}^{n} = {x}^{m+n} \: \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{m} \div {x}^{n} = {x}^{m - n} \: \: }} \\ [/tex]
Additional Information :-
[tex]\boxed{\rm{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}) \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})\: }} \\ [/tex]
Step-by-step explanation:
Given: The given expression is-
[tex] \\ \tt\[ \frac{a-a^{-2}}{\sqrt{a}-\sqrt{a^{-1}}}-\frac{2}{\sqrt{a^{3}}}-\frac{1-a^{-2}}{\sqrt{a}+\sqrt{a^{-1}}} \][/tex]
To find: The goal is to simplify the expression.
[tex]\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}[/tex]
Consider the given expression:
It can be expressed as:
[tex] \\ \\ \color{orangered} \[ \begin{array}{l} \tt =\dfrac{a-\frac{1}{a^{2}}}{\sqrt{a}-\sqrt{\frac{1}{a}}}-\dfrac{2}{\sqrt{a^{3}}}-\dfrac{1-\frac{1}{a^{2}}}{\sqrt{a}+\sqrt{\frac{1}{a}}} \\ \\ \\ \tt=\dfrac{\frac{a^{3}-1}{a^{2}}}{\frac{a-1}{\sqrt{a}}}-\dfrac{2 }{\sqrt{a^{3}}}-\dfrac{\frac{a^{2}-1}{a^{2}}}{\frac{a+1}{\sqrt{a}}} \\ \\ \\ \\ \tt =\dfrac{\sqrt{a}\left(a^{3}-1\right)}{a^{2}(a-1)}-\dfrac{2}{\sqrt{a^{3}}}-\dfrac{\sqrt{a}\left(a^{2}-1\right)}{a^{2}(a+1)} \\ \\ \\ \tt =\dfrac{a^{2}+a+1}{a^{\frac{3}{2}}}-\dfrac{2}{a^{\frac{3}{2}}}-\dfrac{a-1}{a^{\frac{3}{2}}} \\ \\ \\ \tt =\dfrac{a^{2}+a+1-2-a+1}{a^{\frac{3}{2}}} \\ \\ \\ \tt=\dfrac{a^{2}}{a^{\frac{3}{2}}} \\ \\ \\ \tt =a^{2-\frac{3}{2}} \\ \\ \\ \tt =\sqrt{a} \end{array} \] [/tex]
Hence, the simplified value is [tex]\tt \sqrt{a}.[/tex]
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Verified answer
Question :- 15
Simplify :-
[tex]\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} } } - \dfrac{2}{ \sqrt{ {a}^{3} } } - \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
[tex]\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} } } - \dfrac{2}{ \sqrt{ {a}^{3} } } - \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } \\ [/tex]
Consider
[tex]\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} }} \\ [/tex]
[tex]\rm \: = \:\dfrac{a - \dfrac{1}{ {a}^{2} } }{ \sqrt{a} - \dfrac{1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \:\dfrac{\dfrac{ {a}^{3} - 1}{ {a}^{2} } }{ \dfrac{a - 1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{3} - 1}{ {a}^{2} } \times \dfrac{ \sqrt{a} }{a - 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{3} - {1}^{3} }{a \times a} \times \dfrac{ \sqrt{a} }{a - 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ \cancel{(a - 1)}( {a}^{2} + 1 + a)}{a \times \sqrt{a} \times \cancel{\sqrt{a} }} \times \dfrac{ \cancel{\sqrt{a}} }{\cancel{a - 1}} \\ [/tex]
[tex]\rm \: = \: \dfrac{{a}^{2} + 1 + a}{ {( \sqrt{a} )}^{2} \times \sqrt{a}} \\ [/tex]
[tex]\rm \: = \: \dfrac{{a}^{2} + 1 + a}{ {( \sqrt{a} )}^{3}} \\ [/tex]
Now, Consider
[tex]\rm \: \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } \\ [/tex]
[tex]\rm \: = \:\dfrac{1 - \dfrac{1}{ {a}^{2} } }{ \sqrt{a} + \dfrac{1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \:\dfrac{\dfrac{ {a}^{2} - 1}{ {a}^{2} } }{ \dfrac{a + 1}{ \sqrt{a} } } \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{2} - 1}{ {a}^{2} } \times \dfrac{ \sqrt{a} }{a + 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{2} - {1}^{2} }{a \times a} \times \dfrac{ \sqrt{a} }{a + 1} \\ [/tex]
[tex]\rm \: = \: \dfrac{ \cancel{(a + 1)}(a - 1)}{a \times \sqrt{a} \times \cancel{\sqrt{a} }} \times \dfrac{ \cancel{\sqrt{a}} }{\cancel{a + 1}} \\ [/tex]
[tex]\rm \: = \: \dfrac{a- 1}{ {( \sqrt{a} )}^{2} \times \sqrt{a}} \\ [/tex]
[tex]\rm \: = \: \dfrac{a- 1}{ {( \sqrt{a} )}^{3}} \\ [/tex]
Now, Consider the given expression, we have
[tex]\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} } } - \dfrac{2}{ \sqrt{ {a}^{3} } } - \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } \\ [/tex]
can be rewritten as
[tex]\rm \: = \: \dfrac{{a}^{2} + 1 + a}{ {( \sqrt{a} )}^{3}} - \dfrac{2}{{( \sqrt{a} )}^{3}} - \dfrac{a - 1}{{( \sqrt{a} )}^{3}} \\ [/tex]
[tex]\rm \: = \: \dfrac{1}{{( \sqrt{a} )}^{3}} \bigg( {a}^{2} + a + 1 - 2 - a + 1 \bigg) [/tex]
[tex]\rm \: = \: \dfrac{1}{{( \sqrt{a} )}^{3}} \bigg( {a}^{2} + \cancel{a} + \cancel{2} - \cancel{2} - \cancel{a} \bigg) [/tex]
[tex]\rm \: = \: \dfrac{ {a}^{2} }{{( \sqrt{a} )}^{3}} [/tex]
[tex]\rm \: = \: \dfrac{{( \sqrt{a} )}^{4}}{{( \sqrt{a} )}^{3}} \\ [/tex]
[tex]\rm \: = \: \sqrt{a} \\ [/tex]
Hence,
[tex]\rm\implies \:\rm \: \dfrac{a - {a}^{ - 2} }{ \sqrt{a} - \sqrt{ {a}^{ - 1} } } - \dfrac{2}{ \sqrt{ {a}^{3} } } - \dfrac{1 - {a}^{ - 2} }{ \sqrt{a} + \sqrt{ {a}^{ - 1} } } = \sqrt{a} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Identities Used :-
[tex]\boxed{\rm{ \:x = \sqrt{x} \times \sqrt{x} \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} +xy + {y}^{2}) \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{m} \times {x}^{n} = {x}^{m+n} \: \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{m} \div {x}^{n} = {x}^{m - n} \: \: }} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
[tex]\boxed{\rm{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}) \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: }} \\ [/tex]
[tex]\boxed{\rm{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})\: }} \\ [/tex]
Step-by-step explanation:
Given: The given expression is-
[tex] \\ \tt\[ \frac{a-a^{-2}}{\sqrt{a}-\sqrt{a^{-1}}}-\frac{2}{\sqrt{a^{3}}}-\frac{1-a^{-2}}{\sqrt{a}+\sqrt{a^{-1}}} \][/tex]
To find: The goal is to simplify the expression.
[tex]\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}[/tex]
Consider the given expression:
[tex] \\ \tt\[ \frac{a-a^{-2}}{\sqrt{a}-\sqrt{a^{-1}}}-\frac{2}{\sqrt{a^{3}}}-\frac{1-a^{-2}}{\sqrt{a}+\sqrt{a^{-1}}} \][/tex]
It can be expressed as:
[tex] \\ \\ \color{orangered} \[ \begin{array}{l} \tt =\dfrac{a-\frac{1}{a^{2}}}{\sqrt{a}-\sqrt{\frac{1}{a}}}-\dfrac{2}{\sqrt{a^{3}}}-\dfrac{1-\frac{1}{a^{2}}}{\sqrt{a}+\sqrt{\frac{1}{a}}} \\ \\ \\ \tt=\dfrac{\frac{a^{3}-1}{a^{2}}}{\frac{a-1}{\sqrt{a}}}-\dfrac{2 }{\sqrt{a^{3}}}-\dfrac{\frac{a^{2}-1}{a^{2}}}{\frac{a+1}{\sqrt{a}}} \\ \\ \\ \\ \tt =\dfrac{\sqrt{a}\left(a^{3}-1\right)}{a^{2}(a-1)}-\dfrac{2}{\sqrt{a^{3}}}-\dfrac{\sqrt{a}\left(a^{2}-1\right)}{a^{2}(a+1)} \\ \\ \\ \tt =\dfrac{a^{2}+a+1}{a^{\frac{3}{2}}}-\dfrac{2}{a^{\frac{3}{2}}}-\dfrac{a-1}{a^{\frac{3}{2}}} \\ \\ \\ \tt =\dfrac{a^{2}+a+1-2-a+1}{a^{\frac{3}{2}}} \\ \\ \\ \tt=\dfrac{a^{2}}{a^{\frac{3}{2}}} \\ \\ \\ \tt =a^{2-\frac{3}{2}} \\ \\ \\ \tt =\sqrt{a} \end{array} \] [/tex]
Hence, the simplified value is [tex]\tt \sqrt{a}.[/tex]