Bài 2:
a: \(3\cdot2^{x+1}-2\cdot2^{x-1}-3>0\)
=>\(6\cdot2^x-2^x-3>0\)
=>\(2^x>\dfrac{3}{5}\)
=>\(x>log_2\left(\dfrac{3}{5}\right)\)
b: \(\left(2+\sqrt{3}\right)^{2x-3}>=\left(2-\sqrt{3}\right)^{x+1}\)
=>2x-3>=x+1
=>x>=4
c: \(\left(\dfrac{3}{4}\right)^{x^2-2}< =\left(\dfrac{3}{4}\right)^x\)
=>\(x^2-2>=x\)
=>\(x^2-x-2>=0\)
=>(x-2)(x+1)>=0
=>\(\left[{}\begin{matrix}x>=2\\x< =-1\end{matrix}\right.\)
Bài 1:
a: \(3^{x+1}>2\)
=>\(x+1>log_32\)
=>\(x>log_32-1\)
b: \(2\cdot5^x< 3\)
=>\(5^x< \dfrac{3}{2}\)
=>\(x< log_5\left(\dfrac{3}{2}\right)\)
c: \(2-3\cdot2^x>=0\)
=>\(3\cdot2^x< =2\)
=>\(2^x< =\dfrac{2}{3}\)
=>\(x< =log_2\left(\dfrac{2}{3}\right)\)
d: \(\left(\dfrac{3}{4}\right)^{x^2-4}>=1\)
=>\(x^2-4< =0\)
=>(x-2)(x+2)<=0
=>-2<=x<=2
e: \(2+3\left(\dfrac{2}{3}\right)^x< =0\)
=>\(3\cdot\left(\dfrac{2}{3}\right)^x< =-2\)
=>\(\left(\dfrac{2}{3}\right)^x< =-\dfrac{2}{3}\)
=>\(x\in\varnothing\)