g: \(x^2-7x+12>=0\)

=>(x-3)(x-4)>=0

TH1: \(\left\{{}\begin{matrix}x-3>=0\\x-4>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x>=4\end{matrix}\right.\Leftrightarrow x>=4\)

TH2: \(\left\{{}\begin{matrix}x-3< =0\\x-4< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =3\\x< =4\end{matrix}\right.\Leftrightarrow x< =3\)

h: \(2x^2+9x+10< =0\)

=>(x+2)(2x+5)<=0

=>\(-\dfrac{5}{2}< =x< =-2\)

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

=>(x+2)(x-1)<=0(Vì \(2x^2+4>=4>0\forall x\))

TH1: \(\left\{{}\begin{matrix}x+2< =0\\x-1>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =-2\\x>=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x+2>=0\\x-1< =0\end{matrix}\right.\Leftrightarrow-2< =x< =1\)

a: (x+2)(x-3)>0

TH1: \(\left\{{}\begin{matrix}x+2>0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x>3\end{matrix}\right.\)

=>x>3

TH2: \(\left\{{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -2\\x< 3\end{matrix}\right.\)

=>x<-2

b: (x+1)(2x+1)>=0

TH1: \(\left\{{}\begin{matrix}x+1>=0\\2x+1>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\x>=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x>=-\dfrac{1}{2}\)

TH2: \(\left\{{}\begin{matrix}x+1< =0\\2x+1< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\x< =-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x< =-1\)

c: (2x-3)(x+5)<0

TH1: \(\left\{{}\begin{matrix}2x-3< 0\\x+5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x>-5\end{matrix}\right.\Leftrightarrow-5< x< \dfrac{3}{2}\)

Th2: \(\left\{{}\begin{matrix}2x-3>0\\x+5< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -5\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

d: ĐKXĐ: \(x\ne\dfrac{3}{2}\)

\(\dfrac{x-4}{2x-3}>0\)

TH1: \(\left\{{}\begin{matrix}x-4>0\\2x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>4\\x>\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow x>4\)

TH2: \(\left\{{}\begin{matrix}x-4< 0\\2x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 4\\x< \dfrac{3}{2}\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{2}\)

e: ĐKXĐ: \(x\ne-1\)

\(\dfrac{x+3}{x+1}< =0\)

TH1: \(\left\{{}\begin{matrix}x+3< =0\\x+1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =-3\\x>-1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x+3>=0\\x+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\x< -1\end{matrix}\right.\Leftrightarrow-3< =x< -1\)

f: ĐKXĐ: x<>-3/4

\(\dfrac{x-1}{4x+3}>=0\)

TH1: \(\left\{{}\begin{matrix}x-1>=0\\4x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x>-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x>=1\)

TH2: \(\left\{{}\begin{matrix}x-1< =0\\4x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =1\\x< =-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x< =-\dfrac{3}{4}\)

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

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

\(d.\dfrac{4}{2x-3}>0\\ < =>2x-3>0\\ < =>2x>3\\ < =>x>\dfrac{3}{2}\\ e.\dfrac{-3}{x+1}\le0\\ < =>x+1>0\\ < =>x>-1\\ f.\dfrac{2x-1}{2x+3}\ge1\\< =>\dfrac{2x+3-4}{2x+1}\ge1\\ < =>1-\dfrac{4}{2x+1}\ge1\\ < =>\dfrac{-4}{2x+1}\ge0\\ < =>2x+1< 0\\ < =>2x< -1\\ < =>x< -\dfrac{1}{2}\)

d.42x−3>0

=>2x−3>0

=>2x>3

=>x>32

e.−3x+1≤0

=>x+1>0

=>x>−1

f.2x−12x+3≥1

=>2x+3−42x+1≥1

=>1−42x+1≥1

=>−42x+1≥0

=>2x+1<0

=>2x<−1

=>x<−12

#ko cần cảm ơn chỉ cân like là được he!

b: \(\left(x+1\right)\left(x^2+x+1\right)>=0\)

mà \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)

nên x+1>=0

=>x>=-1

c: \(\left(2x-3\right)\left(x^2-2x+3\right)< 0\)

mà \(x^2-2x+3=\left(x-1\right)^2+2>=2>0\forall x\)

nên 2x-3<0

=>2x<3

=>\(x< \dfrac{3}{2}\)

\(4)CO_2+C\rightarrow2CO\\ 5)2NO+O_2\rightarrow2NO_2\\ 7)2KNO_3\rightarrow2KNO_2+O_2\\ 8)2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)