Ta có: \(cos\left(3x+\dfrac{\pi}{6}\right)=\dfrac{1}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{6}=arccos\left(\dfrac{1}{5}\right)+k2\pi\\3x+\dfrac{\pi}{6}=-arccos\left(\dfrac{1}{5}\right)+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=arccos\left(\dfrac{1}{5}\right)-\dfrac{\pi}{6}+k2\pi\\3x=-arccos\left(\dfrac{1}{5}\right)-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{arccos\left(\dfrac{1}{5}\right)}{3}-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=-\dfrac{arccos\left(\dfrac{1}{5}\right)}{3}-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

Điện trở dây dẫn là: \(R=\rho\dfrac{l}{S}=1,1.10^{-6}.\dfrac{15}{0,3.10^{-6}}=5,5\Omega\)

Cường độ dòng điện qua dây dẫn là: \(I=\dfrac{U}{R}=\dfrac{220}{5,5}=40\left(A\right)\)

Hi, cho mình hỏi tí là câu b \(R_5\) không có trong mạch hả?

Ta có: \(8< 9\Leftrightarrow\sqrt{8}< \sqrt{9}\Rightarrow2\sqrt{2}< 3\Rightarrow-2\sqrt{2}>-3\)

Có: \(4-2\sqrt{2}>4-3\Leftrightarrow4-2\sqrt{2}>1\)

\(\Rightarrow\dfrac{4-2\sqrt{2}}{2}>\dfrac{1}{2}\) hay \(2-2\sqrt{2}>\dfrac{1}{2}\)

b. \(\dfrac{2x+4}{4x^2-x}:\dfrac{x^2+2x}{1-4x}\) \(\left(x\ne-2;0;\dfrac{1}{4}\right)\)

\(=\dfrac{2\left(x+2\right)}{x\left(4x-1\right)}.\dfrac{-\left(4x-1\right)}{x\left(x+2\right)}\)

\(=\dfrac{-2}{x^2}\)