Câu hỏi của Trần Tiến Đạt

Question

Tìm n biết:a) $\dfrac{32}{\left(-2\right)^n}=4$b) $\dfrac{8}{2^n}$$=2$c) $\left(\dfrac{1}{2}\right)^{2n-1}$$=\dfrac{1}{8}$

Trúc Giang · Accepted Answer

a) $\dfrac{32}{\left(-2\right)^n}=4$$\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3$=> n = 3b) $\dfrac{8}{2^n}=2$$\Rightarrow2^n=4=2^2$=> n = 2c) $\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}$$\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3$=> 2n - 1 = 3=> 2n = 4=> n = 2Câu trả lời: a) $\dfrac{32}{\left(-2\right)^n}=4$$\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3$=> n = 3b) $\dfrac{8}{2^n}=2$$\Rightarrow2^n=4=2^2$=> n = 2c) $\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}$$\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3$=> 2n - 1 = 3=> 2n = 4=> n = 2

꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜ... · Answer

Giải:a) $\dfrac{32}{\left(-2\right)^n}=4$ $\Rightarrow\left(-2\right)^n=32:4=8$ $\Rightarrow\left(-2\right)^n=8$ Vì $\left(-2\right)^n=2^3$ là ko thể nên n ∈ ∅b) $\dfrac{8}{2^n}=2$$\Rightarrow2^n=8:2=4$ $\Rightarrow2^n=4$ $\Rightarrow2^n=2^2$ $\Rightarrow n=2$ c) $\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}$ $\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3$ $\Rightarrow2n-1=3\rightarrow n=2$Câu trả lời: Giải:a) $\dfrac{32}{\left(-2\right)^n}=4$ $\Rightarrow\left(-2\right)^n=32:4=8$ $\Rightarrow\left(-2\right)^n=8$ Vì $\left(-2\right)^n=2^3$ là ko thể nên n ∈ ∅b) $\dfrac{8}{2^n}=2$$\Rightarrow2^n=8:2=4$ $\Rightarrow2^n=4$ $\Rightarrow2^n=2^2$ $\Rightarrow n=2$ c) $\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}$ $\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3$ $\Rightarrow2n-1=3\rightarrow n=2$

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