a) 2^-1.2^x + 4.2^x = 9.2⁵

=> 2^x-1 + 4.2.2^x-1 = 9.2⁵

=> 2^x-1 + 8.2^x-1 = 9.2⁵

=> 2^x-1.(1 + 8) = 9.2⁵

=> 2^x-1.9 = 9.2⁵

=> x - 1 = 5

=> x = 5 + 1 = 6

Vậy x = 6

b) (7x + 2)^-1 = \(\frac{1}{9}\)

=> 7x + 2 = 9

=> 7x = 9 - 2

=> 7x = 7

=> x = 7 : 7 = 1

Vậy x = 1

a)     2^(-1) . 2^x + 4 . 2^x = 9 . 2^5

=>    1/2 . 2^x + 4 . 2^x     = 9 . 32

=>    2^x . (1/2 + 4)           = 288

=>   2^x . 9/2 = 288 

=> 2^x = 288 : 9/2

=> 2^x = 64

      2^x = 2^6

=> x = 6

b) (7x + 2)^(-1) = 1/9

=> 1 phần 7x + 2  = 1/9

=> 7x + 2 = 9

=> 7x = 9 - 2 

=> 7x = 7

=> x = 7 : 7

=> x = 1

b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)

c) \(\left(3x+2\right)^3=-27\)

\(\Rightarrow\left(3x+2\right)^3=\left(-3\right)^3\)

\(\Rightarrow3x+2=-3\)

\(\Rightarrow3x=\left(-3\right)-2\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=\left(-5\right):3\)

\(\Rightarrow x=-\frac{5}{3}\)

Vậy \(x=-\frac{5}{3}.\)

Chúc bạn học tốt!

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a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)

\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)

\(\Rightarrow x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}.\)

d) \(27^x:3^x=9\)

\(\Rightarrow\left(27:3\right)^x=9\)

\(\Rightarrow9^x=9\)

\(\Rightarrow9^x=9^1\)

\(\Rightarrow x=1\)

Vậy \(x=1.\)

e) \(\frac{16}{2^x}=2\)

\(\Rightarrow2^x=16:2\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

m

2^-1.2^x+4.2^x=9.2^5

0,5.2^x+4.2^x=288

2^x.(4+0,5)=288

2^x.4,5=288

2^x=288:4,5

2^x=64

2^x=2^6

=) x=6

\(8;a,3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)

\(=\frac{3^2.\left(3^4\right)^2}{243.3^3}\)

\(=\frac{3^2.3^8}{3^5.3^3}\)

\(=\frac{3^{10}}{3^8}=3^2=9\)

\(b,\frac{4.2^5}{2^3.\frac{1}{16}}\)

\(=\frac{2^2.2^5}{2^3.\frac{1}{2^4}}\)

\(=\frac{2^7}{\frac{1}{2}}=2^7.2=2^8\)

a, \(3^2.\frac{1}{243}.81^2.\frac{1}{3}^3\)

\(=3^2.\frac{1}{243}.\left(3^4\right)^2.\frac{1}{27}\)

\(=3^2.\frac{1}{243}.3^8.\frac{1}{27}\)

\(=\frac{3^2.3^8}{243.27}\)

\(=\frac{3^2.3^8}{3^5.3^3}\)

\(=\frac{3^{10}}{3^8}=3^2=9\)

b, \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)\)

\(=\left(2^2.2^5\right):\left(8.\frac{1}{16}\right)\)

\(=2^7:\frac{1}{2}\)

\(=2^8\)