Xét các khẳng định sau:

a) hàm số ${\lim }_{x\to 2}\left(3x^3-2x+1\right)=3.2^3-2.2+1=21\Rightarrow$ ĐÚNG

b) Hàm số xác định khi $2x-3\ge 0\Rightarrow x\ge \frac{3}{2}\Rightarrow$ hàm số không liên tục tại $x_0=1\Rightarrow$ SAI

c) ${\lim }_{x\to 1}\left[3f(x)-2g(x)+2025\right]=3{\lim }_{x\to 1}f(x)-2{\lim }_{x\to 1}g(x)+{\lim }_{x\to 1}2025=2009\Rightarrow$ ĐÚNG

d) ${\lim }_{x\to -\infty }\left(\sqrt{2025x^2+bx+1}+ax\right)=-\frac{1}{90}\Rightarrow {\lim }_{x\to -\infty }\left(\frac{2025x^2+bx+1-(ax{)}^2}{\sqrt{2025x^2+bx+1}-ax}\right)=-\frac{1}{90}$

$\Rightarrow {\lim }_{x\to -\infty }\left[\frac{(2025-a^2)x^2+x\left(b+\frac{1}{x}\right)}{-x\left(\sqrt{2025+\frac{b}{x}+\frac{1}{x^2}}+a\right)}\right]=-\frac{1}{90}\Rightarrow \{\begin{array}{l}2025-a^2=0\\ \frac{b}{-(45+a)}=-\frac{1}{90}\end{array}(a\ne -45)\Rightarrow \{\begin{array}{l}a=45\\ b=1\end{array}$

$\Rightarrow a+b=46\Rightarrow$ SAI
a) a) hàm số ${\lim }_{x\to 2}\left(3x^3-2x+1\right)=3.2^3-2.2+1=21\Rightarrow$ ĐÚNG
b) b) Hàm số xác định khi $2x-3\ge 0\Rightarrow x\ge \frac{3}{2}\Rightarrow$ hàm số không liên tục tại $x_0=1\Rightarrow$ SAI
c) c) ${\lim }_{x\to 1}\left[3f(x)-2g(x)+2025\right]=3{\lim }_{x\to 1}f(x)-2{\lim }_{x\to 1}g(x)+{\lim }_{x\to 1}2025=2009\Rightarrow$ ĐÚNG
d) d) ${\lim }_{x\to -\infty }\left(\sqrt{2025x^2+bx+1}+ax\right)=-\frac{1}{90}\Rightarrow {\lim }_{x\to -\infty }\left(\frac{2025x^2+bx+1-(ax{)}^2}{\sqrt{2025x^2+bx+1}-ax}\right)=-\frac{1}{90}$

$\Rightarrow {\lim }_{x\to -\infty }\left[\frac{(2025-a^2)x^2+x\left(b+\frac{1}{x}\right)}{-x\left(\sqrt{2025+\frac{b}{x}+\frac{1}{x^2}}+a\right)}\right]=-\frac{1}{90}\Rightarrow \{\begin{array}{l}2025-a^2=0\\ \frac{b}{-(45+a)}=-\frac{1}{90}\end{array}(a\ne -45)\Rightarrow \{\begin{array}{l}a=45\\ b=1\end{array}$

$\Rightarrow a+b=46\Rightarrow$ SAI