27<3x<3.81 giúp mik với
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\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Rightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\\left[{}\begin{matrix}2x=2\\2x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)


1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)
\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)
\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)
\(\Rightarrow3^x=3\)
\(\Rightarrow x=1\)
2) \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)
\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)
\(\Rightarrow2^{4-x}=2^{50}\)
\(\Rightarrow4-x=50\)
\(\Rightarrow x=-46\)
3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)
\(\Rightarrow2^{2x-3}=2^{49}\)
\(\Rightarrow2x-3=49\)
\(\Rightarrow2x=52\)
\(\Rightarrow x=26\)


-18<x<17
\(\Rightarrow x\in\left\{-18;-17;-16;...;17\right\}\)
-27<x<27
\(\Rightarrow x\in\left\{-27;-26;-25;...;27\right\}\)
|x|<3
\(\Rightarrow x\in\left\{\pm1;\pm5\right\}\)
|-x|<5
\(\Rightarrow x\in\left\{\pm1\right\}\)
chúc bạn học tốt !!!

a, \(27< 3^x< 3\cdot81\)
=> \(3^3< 3^x< 3\cdot3^4\)
=> \(3^3< 3^x< 3^5\)
=> x = 4
b, \(4^{15}\cdot9^{15}< 2^x\cdot3^x< 18^{16}\cdot216\)
=> \(\left[2^2\right]^{15}\cdot\left[3^2\right]^{15}< 2^x\cdot3^x< \left[2\cdot3^2\right]^{16}\cdot6^3\)
=> \(2^{30}\cdot3^{30}< 2^x\cdot3^x< 2^{16}\cdot3^{32}\cdot2^3\cdot3^3\)
=> \(2^{30}\cdot3^{30}< 2^x\cdot3^x< 2^{19}\cdot3^{35}\)
Đến đây tìm được x
\(c,2^{x+1}\cdot3^y=2^{2x}\cdot3^x\Leftrightarrow\frac{2^{2x}}{2^{x+1}}=\frac{3^y}{3^x}\Leftrightarrow2^{x-1}=3^{y-x}\)
\(\Leftrightarrow x-1=y-x=0\Leftrightarrow x=1\)
\(d,6^x:2^{2000}=3^y\)
=> \(\frac{6^x}{3^y}=2^{2000}\)
=> \(\frac{3^{2x}}{3^y}=2^{2000}\)
=> \(3^{2x-y}=2^{2000}\)
Đến đây tìm thử x,y


-20 < x < 21
=> x E {-19;-18;-17;...;19;20}
tổng các số nguyên x là:
(-19)+(-18)+(-17)+...+19+20
= [(-19)+19] + [(-18)+18] + [(-17)+17] +...+ [(-1)+1] + 20 + 0
= 0 + 0 + 0 +...+ 0 + 0 +20
= 20
2 câu sau bạn làm tương tự ( nhớ cộng 2 số đối nhau để tính nhanh hơn )
lxl <_ 3
TH1 TH2
l-xl <_ l3l lxl <_ l3l
-x <_ 3 x <_ 3
=> -x E {0;1;2;3} => x E {0;1;2;3}
=> x E {0;-1;-2;-3}
Vậy x E {0;1;-1;2;-2;3;-3}
l -x l <5
TH1 TH2
l-(-x)l < l5l l-xl < l5l
lxl < l5l -x < 5
x < 5 => -x E {0;1;2;3;4;5}
=>x E {0;1;2;3;4;5} => x E {0;-1;-2;-3;-4;-5}
vậy x E {0;1;-1;2;-2;3;-3;4;-4;5;-5}
ta có :
\(3^3=27< 3^x< 3^5=3.81\)
Vậy \(3< x< 5\)