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Nên M trùng N
MCD:R1nt(R2//R4)nt(R3//R5)
a,\(R_{24}=\dfrac{R_2\cdot R_4}{R_2+R_4}=\dfrac{4\cdot5}{4+5}=\dfrac{20}{9}\left(\Omega\right)\)
\(R_{35}=\dfrac{R_3\cdot R_5}{R_3+R_5}=\dfrac{6\cdot10}{6+10}=3,75\left(\Omega\right)\)
\(R_{tđ}=R_1+R_{24}+R_{35}=2+\dfrac{20}{9}+3,75=\dfrac{287}{36}\left(\Omega\right)\)
\(I_1=I_{24}=I_{35}=I=\dfrac{U}{R_{tđ}}=\dfrac{40}{\dfrac{287}{36}}=\dfrac{1440}{287}\left(A\right)\)
\(U_2=U_4=U_{24}=I_{24}\cdot R_{24}=\dfrac{1440}{287}\cdot\dfrac{20}{9}=\dfrac{3200}{287}\left(V\right)\)
\(U_3=U_5=U_{35}=I_{35}\cdot R_{35}=\dfrac{1440}{287}\cdot3,75=\dfrac{5400}{287}\left(V\right)\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{\dfrac{3200}{287}}{4}=\dfrac{800}{287}\left(A\right)\)
\(I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{5400}{287}}{6}=\dfrac{900}{287}\left(A\right)\)
\(I_4=\dfrac{U_4}{R_4}=\dfrac{\dfrac{3200}{287}}{5}=\dfrac{640}{287}\left(A\right)\)
\(I_5=\dfrac{U_5}{R_5}=\dfrac{\dfrac{5400}{287}}{10}=\dfrac{540}{287}\left(A\right)\)
\(U_1+U_2+U_{MN}+U_5=U\Leftrightarrow R_1I_1+U_2+U_{MN}+U_5=U\)
\(\Rightarrow2\cdot\dfrac{1440}{287}+\dfrac{3200}{287}+U_{MN}+\dfrac{3200}{287}=40\Leftrightarrow U_{MN}=\dfrac{2200}{287}\left(V\right)\)
R1nt(R2//R3)
a) \(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=2\left(\Omega\right)\)
\(\rightarrow R_{td}=R_1+R_{23}=4+2=6\left(\Omega\right)\)
b) Ta có : \(I_1=I_{23}=I=\dfrac{U}{R_{tđ}}=\dfrac{6}{2}=3A\)
\(U_{23}=U_2=U_3=I_{23}.R_{23}=3.2=6V\)
\(\rightarrow I_2=\dfrac{U_2}{R_2}=\dfrac{6}{6}=1A\)
hình ạ 
Ta có: \(U_{AB}=U_{AC}+U_{CD}+U_{DB}=2I_1+5I_2+6\left(I-I_1+I_2\right)\)
\(\Leftrightarrow-4I_1+11I_2+6I=6\left(V\right)^{\left(1\right)}\)
Lại có: \(U_{AB}=U_{AC}+U_{CB}=2I_1+2\left(I_1-I_2\right)\)
\(\Leftrightarrow4I_1-2I_2=6\left(V\right)^{\left(2\right)}\)
Mặt khác: \(U_{AB}=U_{AD}+U_{DB}=3\left(I-I_1\right)+6\left(I-I_1+I_2\right)\Leftrightarrow-9I_1++6I_2+9I=6\left(V\right)^{\left(3\right)}\)
(1)(2)(3) Suy ra:
I1;I2;I
rồi sau đó very ez
@nguyen thi vang @Nguyễn Hải Dương @Ma Đức Minh vào đây chơi nè !