Question

Answer 1

g: \(\dfrac{2x-1}{4x^2+3}>=0\)

mà \(4x^2+3>=3>0\forall x\)

nên 2x-1>=0

=>2x>=1

=>\(x>=\dfrac{1}{2}\)

h:

ĐKXĐ: \(x\ne\dfrac{1}{2}\)

 \(\dfrac{-2025\left(x^2+1\right)}{2x-1}>=0\)

mà -2025<0

nên \(\dfrac{x^2+1}{2x-1}< =0\)

mà \(x^2+1>=1>0\)

nên 2x-1<0

=>2x<1

=>\(x< \dfrac{1}{2}\)

i: \(\left(x^2-x+2\right)\left(2x^2+4\right)\left(3x-1\right)< =0\)

mà \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}>0\forall x;2x^2+4>=4>0\forall x\)

nên 3x-1<=0

=>3x<=1

=>\(x< =\dfrac{1}{3}\)