1 C

2 D

3 C

4 A

5 C

6 A

7 B

8 D

9 A

10 B

11 C

12 B

13 B

14 A

15 B

16 C

11 D

12 B

13 C

14 D

15 C

16 B

Câu 16:

a: =>3x+6=2x+1

=>x=-5

b: =>6(x+3)=7(x-5)

=>7x-35=6x+18

=>x=53

Câu 15 : \(P=tan^3a+cot^3a+tan^2a+cot^2a=\left(tana+cota\right)^3-3tana.cota\left(tana+cota\right)+\left(tana+cota\right)^2-2tana.cota\)

\(=3^3-3.3+3^2-2=25\) . Chọn A 

Câu 16 : Đặt a = \(sin^2x;b=cos^2x\Rightarrow a+b=1\)

Ta có :  \(3a^2+b^2=\dfrac{3}{4}\) \(\Leftrightarrow3a^2+\left(1-a\right)^2=\dfrac{3}{4}\Leftrightarrow4a^2-2a+\dfrac{1}{4}=0\Leftrightarrow\left(a-\dfrac{1}{4}\right)^2=0\Leftrightarrow a=\dfrac{1}{4}\) 

Suy ra : \(a^2=\dfrac{1}{16}\Rightarrow b^2=\dfrac{9}{16}\)

\(P=sin^4x+3cos^4x=a^2+3b^2=\dfrac{1}{16}+3.\dfrac{9}{16}=\dfrac{7}{4}\)  . Chọn C 

Câu 16:

\(Ta.có:n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right);n_{H_2O}=\dfrac{11,7}{18}=0,65\left(mol\right)\\ Vì:n_{H_2O}>n_{CO_2}\Rightarrow Hc:Ankan\)

Đặt 2 hidrocacbon cần tìm có công thức chung là \(C_nH_{2n+2}\left(n>1\right)\)

Ta có: 

\(n_C=n_{CO_2}=0,4\left(mol\right)\\ n_{hh}=n_{H_2O}-n_{CO_2}=0,65-0,4=0,25\left(mol\right)\\ \Rightarrow1< n=\dfrac{0,4}{0,25}=1,6< 2\\ \Rightarrow CTPT.hh:\left\{{}\begin{matrix}CH_4\\C_2H_6\end{matrix}\right.\)

11)

Gọi số mol C₂H₆, C₃H₈ là a, b (mol)

=> \(a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

m_{bình 2 tăng} = m_CO2 

=> \(n_{CO_2}=\dfrac{22}{44}=0,5\left(mol\right)\)

Bảo toàn C: 2a + 3b = 0,5 

=> a = 0,1; b = 0,1

Bảo toàn H: \(n_{H_2O}=\dfrac{0,1.6+0,1.8}{2}=0,7\left(mol\right)\)

=> \(m_{H_2O}=0,7.18=12,6\left(g\right)\)

b) \(\%V_{C_2H_6}=\%V_{C_3H_8}=\dfrac{0,1}{0,2}.100\%=50\%\)

12) 

\(n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\)

Bảo toàn C: n_C(A) = 0,04 (mol)

Số nguyên tử C của A = \(\dfrac{0,04}{0,01}=4\) (nguyên tử)

=> A có CTPT là: C₄H₁₀ 

b) B có CTPT là C₅H₁₂

CTCT: \(CH_3-CH\left(CH_3\right)-CH_2-CH_3\)

Bài 14: 

\(Gọi.chung:C_mH_{2m+2}\left(m>1\right)\\ Ta.có:M_X=16,6.4=66,4\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow14m+2=66,4\\ \Leftrightarrow m=4,6\\ Ta.có:4< m< 5\\ \Rightarrow CTPT:C_4H_{10},C_5H_{12}\\ Đặt:\%n_{C_4H_{10}}=\%V_{C_4H_{10}}=a\\ \Rightarrow\dfrac{58.a+72.\left(100\%-a\right)}{100\%}=66,4\\ \Leftrightarrow a=40\%\\ \Rightarrow\%V_{C_4H_{10}}=40\%;\%V_{C_5H_{12}}=100\%-40\%=60\%\)

1 academy

2 independently

3 educational

4 Generally

5 compulsory

6 activities

7 advanced

8 awarded

9 educative

10 academic 

11 behavior

12 attendance

13 considerably

14 computerise 

15 analysis

16 compulsorily

14: \(=\dfrac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\)