Đề là thế này phải không bạn:\(\left(2x+1\right)\left(x-3\right)=0\)

Ta có:\(\left(2x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=3\end{cases}}\)

Vậy ...........

hỏi chấm

`2x*(-x)=-2x^2`

2x.x=\(2x^2\)

2x^2

\(\left(2x+1\right)^3=-0,001\)

\(\Rightarrow\left(2x+1\right)^3=\left(-0,1\right)^3\)

\(\Rightarrow2x+1=-0,1\)

\(\Rightarrow2x=-0,1-1\)

\(\Rightarrow2x=-1,1\)

\(\Rightarrow x=\dfrac{-1,1}{2}\)

\(\Rightarrow x=-0,55\)

\(G=cot^2x-sin^2x.cot^2x+1-cot^2x=1-sin^2x.cot^2x\)

\(=1-sin^2x.\dfrac{cos^2x}{sin^2x}=1-cos^2x=sin^2x\)

2.

\(tana+cota=2\Rightarrow\left(tana+cota\right)^2=4\)

\(\Rightarrow tan^2a+cot^2a+2tana.cota=4\)

\(\Rightarrow tan^2a+cot^2a+2=4\)

\(\Rightarrow tan^2a+cot^2a=2\)

Ta có: \(\left|2x\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x+3\left(x\ge0\right)\\-2x=x+3\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3\\-2x-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={3;-1}

Chọn C

Đáp án C.