Câu 1: \(x\sqrt{\dfrac{y^2}{x}}=\sqrt{\dfrac{y^2}{x^3}}\)

31) 

BTNT C: \(n_C=n_{CO_2}=n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)\)

Ta có: \(m_{gi\text{ảm}}=m_{BaCO_3}-m_{CO_2}-m_{H_2O}\)

\(\Rightarrow59,1-0,3.44-m_{H_2O}=38,7\Leftrightarrow m_{H_2O}=7,2\left(g\right)\\ \Rightarrow n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\Rightarrow n_H=2n_{H_2O}=0,8\left(mol\right)\)

Ta có: \(n_{H_2O}>n_{CO_2}\left(0,4>0,3\right)\Rightarrow X\) thuộc dãy đồng đẳng ankan

Đặt CTPT của X là \(C_nH_{2n+2}\left(n\in N;n\ge1\right)\)

Ta có: \(n_X=n_{H_2O}-n_{CO_2}=0,1\left(mol\right)\)

Mà \(m_X=0,3.12+0,8=4,4\left(g\right)\)

\(\Rightarrow M_X=\dfrac{4,4}{0,1}=44\left(g/mol\right)\\ \Rightarrow14n+2=44\Leftrightarrow n=3\left(t/m\right)\)

Vậy X là C₃H₈ => Chọn C

c a c a b

C A C A B

\(\left\{{}\begin{matrix}\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=0\\\overrightarrow{G_0B}+\overrightarrow{G_0C}+\overrightarrow{G_0D}=0\end{matrix}\right.\)

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}\)

\(\Leftrightarrow\left(\overrightarrow{GG_0}+\overrightarrow{G_0A}\right)+\left(\overrightarrow{GG_0}+\overrightarrow{G_0B}\right)+\left(\overrightarrow{GG_0}+\overrightarrow{G_0C}\right)+\left(\overrightarrow{GG_0}+\overrightarrow{G_0D}\right)=\overrightarrow{0}\)

\(\Leftrightarrow4\overrightarrow{GG_0}+\overrightarrow{G_0A}+\left(\overrightarrow{G_0B}+\overrightarrow{G_0C}+\overrightarrow{G_0D}\right)=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{AG_0}=4\overrightarrow{GG_0}\)

\(\Rightarrow\overrightarrow{AG}+\overrightarrow{GG_0}=4\overrightarrow{GG_0}\)

\(\Rightarrow\overrightarrow{AG}=3\overrightarrow{GG_0}\)

\(\Rightarrow\overrightarrow{GA}=-3\overrightarrow{GG_0}\)

Câu 31: 

PT: \(CH_2=CH_2+H_2O\underrightarrow{t^o,xt}CH_3CH_2OH\)

→ Đáp án: A

Câu 32:

Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_X=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Gọi CTPT của X là C_nH_2n.

\(\Rightarrow n=\dfrac{n_{CO_2}}{n_X}=5\)

Vậy: X là C₅H_10.

→ Đáp án: D

30.

\(V_{\left(O;\dfrac{3}{4}\right)}\left(M\right)=M_1\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{3}{4}.4=3\\y_1=\dfrac{3}{4}.1=\dfrac{3}{4}\end{matrix}\right.\)

\(Q_{\left(O;-90^0\right)}\left(M_1\right)=M'\Rightarrow\left\{{}\begin{matrix}x'=y_1=\dfrac{3}{4}\\y'=-x_1=-3\end{matrix}\right.\)

\(\Rightarrow M'\left(\dfrac{3}{4};-3\right)\)

31.

\(cos\dfrac{x}{2}=-\dfrac{\sqrt{3}}{2}\Rightarrow\dfrac{x}{2}=\pm\dfrac{5\pi}{6}+k2\pi\)

\(\Rightarrow x=\pm\dfrac{5\pi}{3}+k4\pi\)

Nghiệm dương nhỏ nhất \(x=\dfrac{5\pi}{3}\Rightarrow a+b=5+3=8\)

SO SÁNH À?

e)<

d)<

c)>

b)<

a) lười làm :3