Question

prove that 2^30+2^29+2^28/2^31+2^30-2^29=7/10

Answer 1

2^28(2^2+2+1)/2^28(2^3+2^2-2)
#cancel 2^28
(2^2+2+1)/(2^3+2^2-2)
#simplify
(4+2+1)/(8+4-2)
#simplify
7/10=7/10
#hence proved

Answer 2

Step-by-step explanation:

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

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

$\bf \underline{Solution-}$

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

$\textsf{Taking 2²⁸ Common }$

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

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

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

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

$\textsf{LHS = RHS}$

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