\(M=\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{2001}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(M=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{4002}+\dfrac{11}{4002}\right)\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(M=\left(\dfrac{1}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right)\)

\(M=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)

\(M=1:5\)

\(M=\dfrac{1}{5}\)

\(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(=\left(\dfrac{1}{34}+\dfrac{33}{34}\right):\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)

=1/5

33. B là khẳng định sai

Câu này xác định rất đơn giản, AB và AC cùng thuộc mp (ABC), trong khi MN không song song và cũng ko thuộc (ABC) nên 3 vecto này ko thể đồng phẳng

34.

\(\overrightarrow{MN}=\overrightarrow{MD'}+\overrightarrow{D'C'}+\overrightarrow{C'N}=\dfrac{1}{3}\overrightarrow{AD'}+\overrightarrow{D'C'}+\dfrac{1}{3}\overrightarrow{C'D}\)

\(=\dfrac{1}{3}\overrightarrow{AD}+\dfrac{1}{3}\overrightarrow{AA'}+\overrightarrow{D'C'}+\dfrac{1}{3}\overrightarrow{C'C}+\dfrac{1}{3}\overrightarrow{C'D'}\)

\(=\dfrac{1}{3}\overrightarrow{AD}+\dfrac{2}{3}\overrightarrow{D'C'}=-\dfrac{1}{3}\overrightarrow{AD}+\dfrac{2}{3}\left(B'C'+\overrightarrow{A'B'}\right)\)

\(=-\dfrac{1}{3}\overrightarrow{AD}+\dfrac{2}{3}\overrightarrow{A'C'}\)

A là đáp án đúng

Câu 33:

\(n_{C_6H_2Br_3\left(OH\right)}=\dfrac{4,965}{331}=0,015\left(mol\right)\)

PT: \(C_6H_5OH+3Br_2\rightarrow C_6H_2Br_3\left(OH\right)+3HBr\)

Theo PT: \(n_{Br_2}=3n_{C_6H_2Br_3\left(OH\right)}=0,045\left(mol\right)\)

\(\Rightarrow m_{Br_2}=0,045.160=7,2\left(g\right)\)

Đáp án: D

Câu 34:

Ta có: 94n_C6H5OH + 32n_CH3OH = 15,8 (1)

PT: \(C_6H_5OH+Na\rightarrow C_6H_5ONa+\dfrac{1}{2}H_2\)

\(CH_3OH+Na\rightarrow CH_3ONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_6H_5ONa}+\dfrac{1}{2}n_{CH_3OH}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_6H_5OH}=0,1\left(mol\right)\\n_{CH_3OH}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%n_{CH_3OH}=\dfrac{0,2}{0,1+0,2}.100\%\approx66,67\%\)

Đáp án: C

10: B

33: B

34: D

35: D

33B

\(D=1+3+3^2+3^3+3^4+...+3^{2022}\)

\(3D=3.\left(1+3+3^2+3^3+3^4+...+3^{2022}\right)\)

\(3D=3+3^2+3^3+3^4+3^5+...+3^{2023}\)

\(3D-D=\left(3+3^2+3^3+3^4+3^5+...+3^{2023}\right)-\left(1+3+3^2+3^3+3^4+...+3^{2022}\right)\)

\(2D=\left(3^{2023}-1\right)\)

\(D=\left(3^{2023}-1\right):2\)

3D=3+3^2+...+3^2023

=>2D=3^2023-1

=>\(D=\dfrac{3^{2023}-1}{2}\)

Câu 31: A

Câu 32: C

Câu 33: D

Câu 34: B

Câu 35: C