Toán Lớp 7: $(^{\frac{7}{2^{9}} -\frac{14}{2^{11}}+\frac{21}{768} } )^{2}$ : $( $\frac{5}{2^{9}}$ -$\frac{20}{2^{12}}$+$\frac{25}{1280}$ )^{2}$
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Toán Lớp 7: $(^{\frac{7}{2^{9}} -\frac{14}{2^{11}}+\frac{21}{768} } )^{2}$ : $( $\frac{5}{2^{9}}$ -$\frac{20}{2^{12}}$+$\frac{25}{1280}$ )^{2}$
Comments ( 1 )
Giải đáp:
$\dfrac{49}{25}$.
Lời giải và giải thích chi tiết:
$\dfrac{\left(\dfrac7{2^9}-\dfrac{14}{2^{11}}+\dfrac{21}{768}\right)^2}{\left(\dfrac5{2^9}-\dfrac{20}{2^{12}}+\dfrac{25}{1280}\right)^2}\\=\dfrac{\left(\dfrac7{2^9}-\dfrac{2.7}{2^{11}}+\dfrac{3.7}{3.2^8}\right)^2}{\left(\dfrac5{2^9}-\dfrac{2^2.5}{2^{12}}+\dfrac{5.5}{5.2^8}\right)^2}\\=\dfrac{\left(\dfrac7{2^9}-\dfrac7{2^{10}}+\dfrac7{2^8}\right)^2}{\left(\dfrac5{2^9}-\dfrac5{2^{10}}+\dfrac5{2^8}\right)^2}\\=\left(\dfrac{\dfrac7{2^9}-\dfrac7{2^{10}}-\dfrac7{2^8}}{\dfrac5{2^9}-\dfrac5{2^{10}}+\dfrac5{2^8}}\right)^{\large2}\\=\left(\dfrac{7.\!\left(\dfrac1{2^9}-\dfrac1{2^{10}}-\dfrac1{2^8}\right)}{5.\!\left(\dfrac1{2^9}-\dfrac1{2^{10}}+\dfrac1{2^8}\right)}\right)^{\large2}\\=\left(\dfrac75\right)^2\\=\dfrac{49}{25}$