\(8,64\times93+8,64:0,5+8,64:0,25+8,64\)

\(=8,64\times93+8,64\times2+8,64\times4+8,64\times1\)

\(=8,64\times\left(92+4+2+1\right)\)

\(=8,64\times100\)

\(=864\)

a)9200

9200

6400

b)50

50

19,5

c)128

128

60

a)    92 : 0,01 = 9200          92 x 100 = 9200        64 : 0,01= 6400

b)    25 : 0,5 = 50           25 x 2 = 50           39 : 0,5 = 78
c)    32 : 0,25 = 128         32 x 4 = 128          15 : 0,25 = 60

\(\left(0,75\times64-32\times2\times\frac{3}{4}\right):\left(0,25\times19\times2\times2017\right)\)

= \(\left[64\times\left(0,75-\frac{3}{4}\right)\right]:\left(0,25\times19\times2\times2017\right)\)

= \(\left(64\times0\right):\left(0,25\times19\times2\times2017\right)\)

= \(0:\left(0,25\times19\times2\times2017\right)\)

= \(0\)

\(\left[\left(0,5\right)^3\right]^x=\dfrac{1}{64}\\ \Leftrightarrow\left[\left(\dfrac{1}{2}\right)^3\right]^x=\dfrac{1}{64}\\ \Leftrightarrow\left(\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\\ Vậy:x=2\\ \rightarrow S=\left\{2\right\}\)

Ta có: \(\left[\left(-0.5\right)^3\right]^x=\dfrac{1}{64}\)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^{3x}=\dfrac{1}{64}\)

\(\Leftrightarrow3x=6\)

hay x=2

\(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^x=\left(-\dfrac{1}{8}\right)^2\)

\(\Rightarrow x=2\)

Ta có: \(\left[\left(-0.5\right)^3\right]^x=\dfrac{1}{64}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{3x}=\dfrac{1}{64}\)

\(\Leftrightarrow3x=6\)

hay x=2

\(y\text{×}5+y:0,25=81\)

\(y\text{×}5+y\text{×}4=81\)

\(y\text{×}\left(5+4\right)=81\)

\(y\text{×}9=81\)

\(y=81:9\)

\(y=9\)

\(y\text{×}3+y:0,2=64\)

\(y\text{×}3+y\text{×}5=64\)

\(y\text{×}\left(3+5\right)=64\)

\(y\text{×}8=64\)

\(y=64:8\)

\(y=8\)

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) \(\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)

\(\Rightarrow\left(-4\right)^x=-4\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)

\(\Rightarrow x=1\)

\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)

\(b,\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow7^x=-2401:\left(-7\right)\)

\(\Rightarrow7^x=343\)

\(\Rightarrow7^x=7^3\)

\(\Rightarrow x=3\)

\(c,\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)

\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)

\(\Rightarrow x=-1\)

#\(Toru\)

Câu hỏi là gì bn

Tìm x ạ