Question

[phamthiminhanh]

Bài 1: Giải bất phương trình:

a) 3^x+1>2

b) 2.5^x<3

c) 2-3.2^x≥0

d) (3/4)^{x^2-4}≥1

e) 2+3(2/3)^x≤0

Bài 2: Giải phương trình:

a) 3.2^x+1-2.2^x-1-3>0

b) (2+√3)^2x-3≥(2-√3)^x+1

c) (3/4)^{x^2-2}≤(3/4)^x

Answer 1

[Nguyễn Lê Phước Thịnh]

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: \(2+3\left(\dfrac{2}{3}\right)^x< =0\)

=>\(3\cdot\left(\dfrac{2}{3}\right)^x< =-2\)

=>\(\left(\dfrac{2}{3}\right)^x< =-\dfrac{2}{3}\)

=>\(x\in\varnothing\)