Question

prove that:-
${2}^{30} + {2}^{29} + {2}^{28} \div {2}^{31} + {2}^{30} - {2}^{29} = 7 \div 10$
​

Answer 1

Step-by-step explanation:

$\frac{ {2}^{30} + {2}^{29} + {2}^{28} }{ {2}^{31} + {2}^{30} - {2}^{29} } \\ \\ = \frac{ {2}^{28}( {2}^{2} + 2 + 1) }{ {2}^{29} ( {2}^{2} + 2 - 1) } \\ \\ = \frac{4 + 2 + 1}{2(4 + 2 - 1)} \\ \\ = \frac{7}{10}$

Answer 2

Step-by-step explanation:

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

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

$\bf \underline{Solution-}$

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

$\textsf{Taking 2²⁸ Common }$

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

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

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

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

$\textsf{LHS = RHS}$

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