Math, asked by muhammedeshamisawsem, 11 months ago

Prove that 2^30+2^29+2^28/2^31+2^30-2^29= 7/10
I need it step by step and quickly, please

Answers

Answered by jekkoi46
1

Answer:

Step-by-step explanation:

Attachments:
Answered by Salmonpanna2022
1

Step-by-step explanation:

 \bf \underline{Prove\: that-} \\

   \sf\dfrac{ {2}^{30} +  {2}^{29}  +  {2}^{28}  }{ {2}^{31} +  {2}^{30}  -  {2}^{29}  }  =  \dfrac{7}{10}

 \bf \underline{Solution-} \\

{~~~~~~:~~~\implies\sf\dfrac{ {2}^{30} +  {2}^{29}  +  {2}^{28}  }{ {2}^{31} +  {2}^{30}  -  {2}^{29}  }  =  \dfrac{7}{10} }\\

\textsf{Taking 2²⁸ Common }

\\{~~~~~~:~~~\implies\sf\dfrac{  {2}^{28}( {2}^{2} +  {2}^{1}  +  {2}^{0}  )}{  {2}^{28} ({2}^{3} +  {2}^{2}  -  {2}^{1})  }  =  \dfrac{7}{10} }\\

\\{~~~~~~:~~~\implies\sf\dfrac{   \cancel{{2}^{28}}( 4 +  2  +  1  )}{   \cancel{{2}^{28} }(8 +  4 -  2)  }  =  \dfrac{7}{10} }\\

\\{~~~~~~:~~~\implies\sf\dfrac{   1 \times   7}{  1(12 - 2)  }  =  \dfrac{7}{10} }\\

\\{~~~~~~:~~~\implies\sf\dfrac{  7}{  10}  =  \dfrac{7}{10} } \\

\textsf{LHS = RHS}\\

 \bf \underline{Hence\: proved.} \\

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