Question

$1 \sqrt{2} + 1 \sqrt{2} \div \ \\ sqrt{2} + sqrt{2}$
​

Answer 1

Given :

$\implies \sf 1 \sqrt{2} + 1 \sqrt{2} \div \ \\ \sqrt{2} + \sqrt{2}$

Solution :+

$\dashrightarrow\:\:\sf 1 \sqrt{2} + 1 \sqrt{2} \div \ \\ \sqrt{2} + \sqrt{2}$

$\dashrightarrow\:\:\sf 2 \sqrt{2} \div \ \sqrt{2} + \sqrt{2}$

$\dashrightarrow\:\:\sf \dfrac{2 \sqrt{2}}{ 2 \sqrt{2}}$

$\sf \longrightarrow \: {\dfrac{ 2 \cancel{ \sqrt{2}}^{ \: \: 1} }{ 2 \cancel{ \sqrt{2} }^{ \: \: } } \:}$

$\dashrightarrow\:\:\sf 1$

Therefore,

$\dashrightarrow\:\: \underline{ \boxed{\sf Required \: answer = 1 }}$

Answer 2

Appropriate Question :

• Solve : $\longmapsto \:\bf{\dfrac{1 \sqrt{2} + 1 \sqrt{2} }{ \sqrt{2} + \sqrt{2}}}$

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Given : $\longmapsto \:\bf{Expression \:\:= \dfrac{1 \sqrt{2} + 1 \sqrt{2} }{ \sqrt{2} + \sqrt{2}} }$

Need To Solve : $\longmapsto \:\sf{\dfrac{1 \sqrt{2} + 1 \sqrt{2} }{ \sqrt{2} + \sqrt{2}}}$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Solving \: the \: Given \: Expression \::}}$

$\qquad \quad \dag\:\:\longmapsto \:\sf{\dfrac{1 \sqrt{2} + 1 \sqrt{2} }{ \sqrt{2} + \sqrt{2}}}$

$\longmapsto \:\sf{ \dfrac{\purple {1 \sqrt{2} + 1 \sqrt{2}} }{ \sqrt{2} + \sqrt{2}}}$

$\longmapsto \:\sf{\dfrac{\purple {2 \sqrt{2} } }{ \sqrt{2} + \sqrt{2}} }$

$\longmapsto \:\sf{\dfrac{2 \sqrt{2} }{ \sqrt{2} + \sqrt{2}}}$

$\longmapsto \:\sf{\dfrac{2 \sqrt{2} }{ \purple {\sqrt{2} + \sqrt{2}}}}$

$\longmapsto \:\sf{\dfrac{2 \sqrt{2} }{ \purple {2\sqrt{2} }}}$

$\longmapsto \:\sf{\dfrac{2 \sqrt{2} }{ 2\sqrt{2}} }$

$\longmapsto \:\sf{\cancel {\dfrac{2 \sqrt{2} }{ 2\sqrt{2} }}}$

$\longmapsto \:\bf{ 1 \qquad \quad \longrightarrow\:\; AnswEr\:}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm { The\: Value \:of\:Given \:Expression \:is\:\bf{1\: }}}}$

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