Math, asked by user1239042, 4 months ago

answer urgently I will mark brainliest

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Answered by XxMissCutiepiexX
7

\huge\boxed{\bf{\red{\fcolorbox{blue}{white}{Answer}}}}

 \bf \scriptsize{As  \: the \:  given \:  Irrational \:  numbers \:  are  \: not \:  similar \:  , so \:  we  \: reduce } \\  \bf \scriptsize{\:  each \:  irrational  \: number \:  in \:  simplest \:  form. \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }

 \bf \scriptsize{ \:  \: 2 \sqrt{8} = 2 \sqrt{2^{2} \times 2} \:  = \: 2 \times  \sqrt{2^{2} } \times  \sqrt{2} \:  =  \: 2 \times  2 \: \times \sqrt{2}  } \\  \\  \bf \scriptsize{ = 4 \sqrt{2} }

 \bf \scriptsize{And  \:  \:  \: \:  \:  \:  \:  \:  \frac{ \:  \: 3}{ \sqrt{6} }  = \frac{ \:  \: 3}{ \sqrt{2} }  \times  \frac{ \sqrt{2} }{ \sqrt{2} } =  \frac{3}{2} \sqrt{2}    }

 \bf \scriptsize{Here,  \:  \:  \: 4 \sqrt{2} - 2 \sqrt{8} +  \frac{  \: \: 3}{ \sqrt{2} }  \sqrt{2} } \\  \\  \bf \scriptsize{ =  \sqrt{2} \:  \: (4 - 4 +  \frac{3}{2} )} \\  \\  \bf \scriptsize{ =  \frac{3}{2} \sqrt{2}  }

Answered by sagacioux
5

Answer:

As the given Irrational numbers are not similar, so we reduce each irrational number in simplest form.

\begin{gathered} \bf \scriptsize{ \: \: 2 \sqrt{8} = 2 \sqrt{2^{2} \times 2} \: = \: 2 \times \sqrt{2^{2} } \times \sqrt{2} \: = \: 2 \times 2 \: \times \sqrt{2} } \\  \\ \\ \bf \scriptsize{ = 4 \sqrt{2} }\end{gathered}</p><p>\bf \scriptsize{And \: \: \: \: \: \: \: \: \frac{ \: \: 3}{ \sqrt{6} } = \frac{ \: \: 3}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{3}{2} \sqrt{2} } \\ </p><p>\begin{gathered} \bf \scriptsize{Here, \: \: \: 4 \sqrt{2} - 2 \sqrt{8} + \frac{ \: \: 3}{ \sqrt{2} } \sqrt{2} }  \\ \\ \\ \bf \scriptsize{ = \sqrt{2} \: \: (4 - 4 + \frac{3}{2} )}  \\ \\ \\ \bf \scriptsize{ = \frac{3}{2} \sqrt{2} }\end{gathered}

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