a.\(x^2-3x=x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

b.\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

c.\(x^3+x^2+x+1=x^2\left(x+1\right)+\left(x+1\right)=\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\text{ ( vô lí )}\\x=-1\end{cases}}}\Leftrightarrow x=-1\)

d.\(x^2+\sqrt{2015}=0\Leftrightarrow x^2=-\sqrt{2015}< 0\text{ (vô nghiệm)}\)

mk ko bt bạn ah !!

(x+2)*(2x-1)=0

TH1: x+2=0              TH2: 2x-1=0

        x=-2                          2x=1

                                            x=1/2

vậy x= -2 hoặc 1/2

TA CÓ :\(\frac{3x^3-x^2-3x+2021}{3x^{4^{ }}-x^3+3x+2021}\)= \(\frac{\left(x^2-1\right)\left(3x-1\right)+2020}{\left(3x-1\right)\left(x^3+1\right)+2022}\)

THAY x=1/3 vào ta được :\(\frac{\left(3.\frac{1}{3}-1\right)\left(\frac{1}{3^2}-1\right)+2020}{\left(3.\frac{1}{3}-1\right)\left(\frac{1}{3^3}+1\right)+2022}\)= \(\frac{2020}{2022}\)=\(\frac{1010}{1011}\)

 Vậy C=1010/1011