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a)\(\frac{x-2}{16}=\frac{-4}{2-x}\)
\(\Rightarrow\frac{x-2}{16}=\frac{4}{x-2}\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=16.4\)
\(\Rightarrow\left(x-2\right)^2=64\)
Mà ta có: \(64=\left(\pm8\right)^2\)
Suy ra: \(\orbr{\begin{cases}\left(x-2\right)^2=8^2\\\left(x-2\right)^2=\left(-8\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Rightarrow}\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)
Vậy x = 10 hoặc x = -6
b) \(\left(2x+7\right)^2-28=64\)
\(\Rightarrow\left(2x+7\right)^2=64+28=92\)
Mà: \(92=\left(\pm2\sqrt{23}\right)^2\)
Nên \(\orbr{\begin{cases}2x+7=2\sqrt{23}\\2x+7=-2\sqrt{23}\end{cases}\Rightarrow\orbr{\begin{cases}2x=2\sqrt{23}-7\\2x=-2\sqrt{23}-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2\sqrt{23}-7}{2}\\x=-\frac{7+2\sqrt{23}}{2}\end{cases}}}\)
Vậy .....
(Bạn xem lại đề nha, kết quả bài b lẻ quá nhưng cách làm vẫn vậy nha!)
a) \(64^x:16^x=256\)
\(\Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=2^8\)
\(\Rightarrow2^{6x}:2^{4x}=2^8\)
\(\Rightarrow2^{6x-4x}=2^8\)
\(\Rightarrow2^{2x}=2^8\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
b) \(\dfrac{-2401}{7^x}=-7\)
\(\Rightarrow\dfrac{-7^4}{7^x}=-7\)
\(\Rightarrow-7^{4-x}=-7\)
\(\Rightarrow7^{4-x}=7\)
\(\Rightarrow4-x=1\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
c) \(\dfrac{64}{\left(-4\right)^x}=-256\)
\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)
\(\Rightarrow\left(-4\right)^x=-4\)
\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)
\(\Rightarrow x=1\)
\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)
\(b,\dfrac{-2401}{7^x}=-7\)
\(\Rightarrow7^x=-2401:\left(-7\right)\)
\(\Rightarrow7^x=343\)
\(\Rightarrow7^x=7^3\)
\(\Rightarrow x=3\)
\(c,\dfrac{64}{\left(-4\right)^x}=-256\)
\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)
\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)
\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)
\(\Rightarrow x=-1\)
#\(Toru\)
Ta có ; \(\frac{x}{y^2}=16\Rightarrow\frac{x}{y}.\frac{1}{y}=16\)
Mà \(\frac{x}{y}=64\)
Nên : \(64.\frac{1}{y}=16\)
\(\Rightarrow\frac{1}{y}=\frac{16}{64}=\frac{1}{4}\)
=> y = 4
Nên x = 64 x 4
=> x = 256
Vậy x = 256 ; y = 4
\(\frac{64}{\left(-2\right)^x}=\left(-16\right)^2:4^3\)
<=> \(\frac{64}{\left(-2\right)^x}=4\)
<=> \(\frac{64}{\left(-2\right)^x}=\frac{64}{16}\)
<=> (-2)x = 16
<=> x = 4
\(\left(\dfrac{9}{16}\right)^5\cdot x=\left(\dfrac{27}{64}\right)^3\)
\(\Leftrightarrow\left(\dfrac{3}{4}\right)^{10}\cdot x=\left(\dfrac{3}{4}\right)^9\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^9:\left(\dfrac{3}{4}\right)^{10}\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^{-1}\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy x=4/3
(\(\dfrac{9}{16}\))5\(\times\) \(x\) = (\(\dfrac{27}{64}\))3
\(x\) = (\(\dfrac{27}{64}\))3 : (\(\dfrac{9}{16}\))5
\(x\) = (\(\dfrac{3^3}{2^6}\))3: (\(\dfrac{3^2}{2^4}\))5
\(x\) = \(\dfrac{3^9}{2^{18}}\) : \(\dfrac{3^{10}}{2^{20}}\)
\(x\) = \(\dfrac{3^9}{2^{18}}\) \(\times\) \(\dfrac{2^{20}}{3^{10}}\)
\(x\) = \(\dfrac{2^2}{3}\)
\(x\) = \(\dfrac{4}{3}\)
\(\frac{\left(-4\right)^x}{64}=-16\)
\(\Leftrightarrow\left(-4\right)^x=64.16\)
\(\Leftrightarrow\left(-4\right)^x=1024\)
\(\Leftrightarrow4^x=1024\)
\(\Leftrightarrow4^x=4^5\)
\(\Rightarrow x=5\)
\(\frac{\left(-4\right)^x}{64}=-16\)
\(\Leftrightarrow\left(-4\right)^x=64.16\)
\(\Leftrightarrow\left(-4\right)^x=1024\)
\(\Leftrightarrow4^x=1024\)
\(\Leftrightarrow4^x=4^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)