Chọn C

`a)` Vì `O` là trung điểm của `AC;BD`

    `=>{(\vec{OA}=-\vec{OC}),(\vec{OB}=-\vec{OD}):}`

Ta có: `\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}`

    `=-\vec{OC}-\vec{OD}+\vec{OC}+\vec{OD}=0`

`b)` Vì `E` là trung điểm `AD=>\vec{EA}=-\vec{ED}`

Ta có: `\vec{EA}+\vec{EB}+2\vec{EC}`

`=\vec{EA}+\vec{EA}+\vec{AB}+2\vec{ED}+2\vec{DC}`

`=-2\vec{ED}+\vec{AB}+2\vec{ED}+2\vec{AB}=3\vec{AB}`

`c)` Ta có: `\vec{EB}+2\vec{EA}+4\vec{ED}`

`=\vec{EB}-2\vec{ED}+4\vec{ED}`

`=\vec{EB}+2\vec{ED}`

`=\vec{EA}+\vec{AB}+2\vec{ED}`

`=-\vec{ED}+\vec{AB}+2\vec{ED}`

`=\vec{AB}+\vec{EC}+\vec{CD}` (Mà `\vec{AB}=-\vec{CD}`)

`=\vec{EC}`

b

B

B

A

D

C-C

good

good->better

good => better

\(a,\dfrac{1}{2}x=3+2\)

\(\dfrac{1}{2}x=5\)

\(x=5\div\dfrac{1}{2}\)

\(x=10\)

\(b,\dfrac{1}{4}x^2-\sqrt{36}=10\)

\(\dfrac{1}{4}x^2-6=10\)

\(\dfrac{1}{4}x^2=10+6\)

\(\dfrac{1}{4}x^2=16\)

\(x^2=16\div\dfrac{1}{4}\)

\(x^2=64\)

\(x^2=\left(8\right)^2\)

\(\Rightarrow x=8\)

\(MCD:\left(R_dntR1\right)//R2\)

\(->R_d=\dfrac{U_d^2}{P_d}=\dfrac{6^2}{3}=12\Omega\)

\(->R_{td}=\dfrac{\left(R_d+R1\right)\cdot R2}{R_d+R1+R2}=\dfrac{\left(12+6\right)\cdot6}{12+6+6}=4,5\Omega\)

\(->I=\dfrac{U}{R}=\dfrac{13,5}{4,5}=3A\)

\(->I_d=I1=\dfrac{P_d}{U_d}=\dfrac{3}{6}=0,5A\)

\(->I2=I-I_d1=3-0,5=2,5A\)

\(I_{AB}=I=3A\)

\(\left\{{}\begin{matrix}P_d=3\\P1=I1^2\cdot R1=0,5^2\cdot6=1,5\\P2=I2^2\cdot R2=2,5^2\cdot6=37,5\\P_{AB}=UI=13,5\cdot3=40,5\end{matrix}\right.\)(W)

Ta có: \(A//R1\)

\(=>U_A=U1=I1\cdot R1=0,5\cdot6=3V\)

\(=>I_A=\dfrac{U_A}{R_A}=\dfrac{3}{0}\) (vô lý)