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

\(\Leftrightarrow4x^2+4x+1-x^2+2x-1=0\)

\(\Leftrightarrow3x^2+6x=0\)

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

Vậy \(S=\left\{0;-2\right\}\).

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

\(\Leftrightarrow9x^2-6x+1-9x^2+12x-4=0\)

\(\Leftrightarrow6x-3=0\)

\(\Leftrightarrow6x=3\)

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

\(x^2=10x-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Kiểm tra lại bước giải nhé.

\(3x^2+12x=0\Leftrightarrow x\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{0;-4\right\}\).

Giả sử : \(x_1< x_2\) thì \(y_1=-5x_1\) ; \(y_2=-5x_2\)

Xét hiệu : \(y_1-y_2=-5x_1-\left(-5x_2\right)\)

                             \(=-5x_1+5x_2\)

                             \(=-5\left(x_1-x_2\right)\)

                             \(=-5\left(x_1-x_2\right)>0\) (vì \(x_1< x_2\))

                             \(\Rightarrow y_1>y_2\)

Vì \(x_1< x_2\) mà \(y_1>y_2\) nên hàm số này là hàm số nghịch biến.

Giả sử \(x_1< x_2\) thì \(y_1=2x_1-3\) ; \(y_2=2x_2-3\)

Xét \(y_1-y_2=\left(2x_1-3\right)-\left(2x_2-3\right)\)

                   \(=2x_1-3-2x_2+3\)

                   \(=2\left(x_1-x_2\right)< 0\) (vì \(x_1< x_2\))

                   \(\Rightarrow y_1< y_2\)

Vậy \(x_1< x_2\) mà \(y_1< y_2\) nên hàm số trên là hàm số đồng biến.

\(AB=13\cos50^\circ\approx8,35\)

\(3x^2+12x=0\) \(\Leftrightarrow x\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(S=\left\{0;-4\right\}\).

1) Ta có: \(\sin\alpha=\dfrac{1}{2}\Rightarrow\alpha=30^\circ\)

\(\Rightarrow\cos\alpha=\dfrac{\sqrt{3}}{2}\)

\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{\sqrt{3}}{3}\)

\(\Rightarrow\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{3}{\sqrt{3}}\)

2) Ta có: \(\cos\alpha=\dfrac{\sqrt{2}}{2}\Rightarrow\alpha=45^{\circ}\)

\(\Rightarrow\sin\alpha=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}=1\)

\(\Rightarrow\cot\alpha=\dfrac{1}{\tan\alpha}=1\)