tìm x thỏa mãn ( $\frac{3}{2x}$ – $\frac{5}{2})^{60}$ = $(\frac{1}{8})^{20}$ : $(-\frac{1}{4})^{10}$ 19/11/2021 Bởi aihong tìm x thỏa mãn ( $\frac{3}{2x}$ – $\frac{5}{2})^{60}$ = $(\frac{1}{8})^{20}$ : $(-\frac{1}{4})^{10}$
Ta có $\left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} = \left( \dfrac{1}{8} \right)^{20} : \left( -\dfrac{1}{4} \right)^{10}$ $<-> \left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} = \dfrac{1}{(2^3)^{20}} : \dfrac{1}{(2^2)^{10}}$ $<-> \left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} =\dfrac{2^{20}}{2^{60}}$ $<-> \left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} = \dfrac{1}{2^{40}}$ $<-> \dfrac{3}{2x} – \dfrac{5}{2} = \pm \dfrac{1}{\sqrt[3]{4}}$ $<-> \dfrac{3}{2x} = \dfrac{5}{2} \pm \dfrac{1}{\sqrt[3]{4}}$ $<-> \dfrac{3}{x} = 5 \pm \sqrt[3]{2}$ $<-> x = \dfrac{3}{5 \pm \sqrt[3]{2}}$ Vậy $x = \dfrac{3}{5 \pm \sqrt[3]{2}}$. Bình luận
Ta có
$\left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} = \left( \dfrac{1}{8} \right)^{20} : \left( -\dfrac{1}{4} \right)^{10}$
$<-> \left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} = \dfrac{1}{(2^3)^{20}} : \dfrac{1}{(2^2)^{10}}$
$<-> \left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} =\dfrac{2^{20}}{2^{60}}$
$<-> \left( \dfrac{3}{2x} – \dfrac{5}{2} \right)^{60} = \dfrac{1}{2^{40}}$
$<-> \dfrac{3}{2x} – \dfrac{5}{2} = \pm \dfrac{1}{\sqrt[3]{4}}$
$<-> \dfrac{3}{2x} = \dfrac{5}{2} \pm \dfrac{1}{\sqrt[3]{4}}$
$<-> \dfrac{3}{x} = 5 \pm \sqrt[3]{2}$
$<-> x = \dfrac{3}{5 \pm \sqrt[3]{2}}$
Vậy $x = \dfrac{3}{5 \pm \sqrt[3]{2}}$.