Em xin cảm ơn cô và mọi người đã dành thời gian đọc bài viết ạ. 

a. Đèn sáng bình thường nên \(I=\dfrac{P_{Đm}}{U_{Đm}}=\dfrac{4,5}{6}=0,75\left(A\right)\)

b. \(A=UIt=Pt=4,5.5=22,5\left(J\right)\)

a. Ta có \(F=k\dfrac{\left|q_1q_2\right|}{r^2}=9.10^9.\dfrac{q_1^2}{r^2}\) \(\Leftrightarrow q_1=\sqrt{\dfrac{r^2.F}{9.10^9}}=\sqrt{\dfrac{0,02^2.10^{-6}}{9.10^9}}\)

\(\Leftrightarrow q_1=q_2=2,1.10^{-10}\left(C\right)\)

b. \(F'=k\dfrac{q_1^2}{r'^2}\Leftrightarrow r'=\sqrt{\dfrac{k.q_1^2}{F'}}=\sqrt{\dfrac{9.10^9.\left(2,1.10^{-10}\right)^2}{5.10^{-6}}}\) \(\Leftrightarrow r'=8,9.10^{-3}\left(m\right)=0,89\left(cm\right)\)

\(Sinx.Cos\left(x-\dfrac{\pi}{4}\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}Sinx=0\\Cos\left(x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{3\pi}{4}+k\pi\end{matrix}\right.\)

Vậy... 

a. \(Q=CU=20.10^{-6}.220=4,4.10^{-3}\left(C\right)\)

B. \(W=\dfrac{Q^2}{2C}=\dfrac{\left(4,4.10^{-3}\right)^2}{2.20.10^{-6}}=0,484\left(J\right)\)

Ta có \(\left\{{}\begin{matrix}e+p+n=28\\e+p-n=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2p+n=28\\2p-n=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}p=e=9\\n=10\end{matrix}\right.\)

\(\dfrac{2x}{x-1}+\dfrac{x-1}{x+1}-\dfrac{3x+1}{x^2-1}\) = \(\dfrac{2x\left(x+1\right)}{x^2-1}+\dfrac{\left(x-1\right)^2}{x^2-1}-\dfrac{3x+1}{x^2-1}\) 

= \(\dfrac{2x^2+2x+x^2-2x+1-3x-1}{x^2-1}\)

\(=\dfrac{3x^2-3x}{x^2-1}\) \(=\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\) \(=\dfrac{3x}{x+1}\)

\(Ca\left(HCO_3\right)_2\rightarrow Ca^{2+}+2HCO_3^-\)

\(HCO_3^-⇌H^++CO_3^{2-}\)