mai mk cx hi hk1 cx thi Cd ti anh do

Câu 1.

\(R_N=R_1+R_2=2+3=5\Omega\)

\(I_1=I_2=I=\dfrac{\xi}{r+R_N}=\dfrac{9}{1+5}=1,5A\)

\(U_N=I\cdot R_N=1,5\cdot5=7,5V\)

\(U_1=I_1\cdot R_1=1,5\cdot2=3V;U_2=7,5-3=4,5V\)

\(P_1=I_1^2\cdot R_1=1,5^2\cdot2=4,5W\)

\(P_2=1,5^2\cdot3=6,75W\)

đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)

Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)

\(=3-\frac{3}{20}=\frac{57}{20}\)

\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)

Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A

\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B

\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)

\(A=\frac{1}{2}.\frac{19}{20}\)

\(A=\frac{19}{40}\)

\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)

\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)

\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)

\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)

\(B=\left[3.\left(\frac{19}{20}\right)\right]\)

\(B=\frac{57}{20}\)

Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)

\(=-\frac{95}{40}=-\frac{19}{8}\)

Nếu đúng thì k nha

hihi mình thì thi rồi, nên không cần phải lo gì nữa hết

Ta có

\(m_{CuCl_2\left(1\right)}=\dfrac{300.15\%}{100\%}=45g\\ C_{\%CuCl_2\left(3\right)}=\dfrac{45+\dfrac{m.5,4\%}{100\%}}{300+m}\cdot100\%=11,8\%\\ \Rightarrow m=150g\)

Chúc thi tốt nhé  

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