ko

2315

13: \(cos^2x-\sqrt3\cdot\sin2x=1+\sin^2x\)

=>\(cos^2x-\sin^2x-\sqrt3\cdot\sin2x=1\)

=>\(cos2x-\sqrt3\cdot\sin2x=1\)

=>\(\frac12\cdot cos2x-\frac{\sqrt3}{2}\cdot\sin2x=\frac12\)

=>\(\sin\left(\frac{\pi}{6}-2x\right)=\frac12\)

=>\(\sin\left(2x-\frac{\pi}{6}\right)=-\frac12\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\ 2x-\frac{\pi}{6}=\pi+\frac{\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=k2\pi\\ 2x=\frac43\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=k\pi\\ x=\frac23\pi+k\pi\end{array}\right.\)

12: \(\sin2x-2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\cdot cosx-2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\left(cosx-\sin x\right)=0\)

=>\(\sin x\cdot\left(\cos x-\sin x\right)=0\)

TH1: sin x=0

=>\(x=k\pi\)

TH2: cos x-sin x=0

=>\(\sin x=cosx\)

=>\(\sin x=\sin\left(\frac{\pi}{2}-x\right)\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{2}-x+k2\pi\\ x=\pi-\frac{\pi}{2}+x+k2\pi=\frac{\pi}{2}+x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{2}+k2\pi\\ \frac{\pi}{2}+k2\pi=0\left(vôlý\right)\end{array}\right.\)

=>\(x=\frac{\pi}{4}+k\pi\)

11: \(\sin2x+2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\cdot cosx+2\cdot\sin^2x=0\)

=>\(2\cdot\sin x\cdot\left(cosx+\sin x\right)=0\)

=>\(\sin x\cdot\left(cosx+\sin x\right)=0\)

TH1: \(\sin x=0\)

=>\(x=k\pi\)

TH2: sin x+cos x=0

=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(x+\frac{\pi}{4}=k\pi\)

=>\(x=k\pi-\frac{\pi}{4}\)

5: \(2\cdot cos^2x-2\cdot cos2x-4\cdot\sin x\cdot cosx+1=0\)

=>\(2\cdot cos^2x-1-2\cdot cos2x-2\cdot\sin2x+2=0\)

=>\(cos2x-2\cdot cos2x-2\cdot\sin2x+2=0\)

=>\(-cos2x-2\cdot\sin2x=-2\)

=>\(2\cdot\sin2x+cos2x=2\)

=>\(\sin2x\cdot\frac{2}{\sqrt5}+cos2x\cdot\frac{1}{\sqrt5}=\frac{2}{\sqrt5}\)

=>\(\sin\left(2x+\alpha\right)=cos\left(\alpha\right)=\sin\left(\frac{\pi}{2}-\alpha\right)\)

=>\(\left[\begin{array}{l}2x+\alpha=\frac{\pi}{2}-\alpha+k2\pi\\ 2x+\alpha=\pi-\frac{\pi}{2}+\alpha+k2\pi=a+\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-2\alpha+\frac{\pi}{2}+k2\pi\\ 2x=\frac{\pi}{2}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\alpha+\frac{\pi}{4}+k\pi\\ x=\frac{\pi}{4}+k\pi\end{array}\right.\)

4: \(2\cdot cos^2x-3\sqrt3\cdot\sin2x-4\cdot\sin^2x+4=0\)

=>\(2\cdot\frac{1+cos2x}{2}-3\sqrt3\cdot\sin2x-4\cdot\frac{1-cos2x}{2}+4=0\)

=>\(1+cos2x-3\sqrt3\cdot\sin2x-2\left(1-cos2x\right)+4=0\)

=>\(1+cos2x-3\sqrt3\cdot\sin2x-2+2\cdot cos2x+4=0\)

=>\(-3\sqrt3\cdot\sin2x+3\cdot cos2x+3=0\)

=>\(-\sqrt3\cdot\sin2x+cos2x=-1\)

=>\(\sqrt3\cdot\sin2x-cos2x=1\)

=>\(\sin2x\cdot\frac{\sqrt3}{2}-cos2x\cdot\frac12=\frac12\)

=>\(\sin\left(2x-\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}2x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ 2x-\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 2x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi\\ x=\frac{\pi}{2}+k\pi\end{array}\right.\)

3: \(4\sqrt3\cdot\sin x\cdot cosx+4\cdot cos^2x-2\cdot\sin^2x=\frac52\)

=>\(2\sqrt3\cdot\sin2x+4\cdot\frac{1+cos2x}{2}-2\cdot\frac{1-cos2x}{2}=\frac52\)

=>\(2\sqrt3\cdot\sin2x+2+2\cdot cos2x-1+cos2x=\frac52\)

=>\(2\sqrt3\cdot\sin2x+3\cdot cos2x=\frac52-1=\frac32\)

=>\(\sin2x\cdot2+cos2x\cdot\sqrt3=\frac{\sqrt3}{2}\)

=>\(\sin2x\cdot\frac{2}{\sqrt7}+cos2x\cdot\frac{\sqrt3}{\sqrt7}=\frac{\sqrt3}{2\sqrt7}\)

=>\(\sin\left(2x+\alpha\right)=\frac{\sqrt3}{2\sqrt7}=\frac{\sqrt{21}}{14}\)

=>\(\left[\begin{array}{l}2x+\alpha=\arcsin\left(\frac{\sqrt{21}}{14}\right)+k\pi\\ 2x+\alpha=\pi-\arcsin\left(\frac{\sqrt{21}}{14}\right)+k\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\arcsin\left(\frac{\sqrt{21}}{14}\right)+k\pi-\alpha\\ 2x=\pi-\arcsin\left(\frac{\sqrt{21}}{14}\right)-\alpha+k\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac12\cdot\arcsin\left(\frac{\sqrt{21}}{14}\right)+\frac{k\pi}{2}-\frac{\alpha}{2}\\ x=\frac{\pi}{2}-\frac12\cdot\arcsin\left(\frac{\sqrt{21}}{14}\right)-\frac{\alpha}{2}+\frac{k\pi}{2}\end{array}\right.\)

13. C

14.C

15.A

ai chỉ mình với, đang cần gấp ạ!

$n_{Fe_3O_4} = 0,08(mol)$

$n_{CO} = 0,3(mol)$

$Fe_3O_4 + 4CO \xrightarrow{t^o} 3Fe + 4CO_2$

Gọi $n_{Fe_3O_4\ pư} =a (mol) \Rightarrow n_{Fe_3O_4\ dư} = 0,08 - a(mol)$

Theo PTHH : $n_{Fe} = 3a(mol)$

Suy ra : $m_{hh\ rắn} = (0,08 - a).232 + 3a.56 = 14,72(gam)$

$\Rightarrow a = 0,06$

Hiệu suất $ = \dfrac{0,06}{0,08}.100\% = 75\%$

1 C

2 D

3 D

4 B

5 B

6 D

7 B

8 A

9 C

10 A

11 B

12 D

13 A

14 B

15 A

\(f'\left(x\right)=\dfrac{1}{cos^2\left(x-\dfrac{2\pi}{3}\right)}\Rightarrow f'\left(0\right)=\dfrac{1}{cos^2\left(-\dfrac{2\pi}{3}\right)}=4\)

Câu 14:

\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\Rightarrow n_H=0,2.2=0,4\left(mol\right)\)

Có: m_C + m_H = 5,2 (g) ⇒ m_C = 5,2 - 0,4.1 = 4,8 (g) 

\(\Rightarrow n_{CO_2}=n_C=\dfrac{4,8}{12}=0,4\left(mol\right)\)

A: C_nH_2n-2

⇒ n_A = n_CO2 - n_H2O = 0,4 - 0,2 = 0,2 (mol)

\(\Rightarrow n=\dfrac{n_{CO_2}}{n_A}=2\)

→ A là C₂H₂.

→ Đáp án: B

Câu 15:

X: C_nH_2n-2

\(\Rightarrow\dfrac{2n-2}{12n+2n-2}=0,11111\Rightarrow n=4\)

→ X là C₄H₆.

CTCT: \(CH\equiv C-CH_2-CH_3\)

\(CH_3-C\equiv C-CH_3\)

→ Đáp án: B

a)  C_M = \(\dfrac{n}{V}\Rightarrow n=C_M.V=0,9.2,5=2,25\left(mol\right)\)

b)                                 250 ml = 0,25 lit

     C_M = \(\dfrac{n}{V}\Rightarrow n=C_M.V=0,1.0,25=0,025\left(mol\right)\)

c) C⁰/₀ = \(\dfrac{m_{ct}.100}{m_{dd}}\Rightarrow m_{ct}=\dfrac{C.m_{dd}}{100}=\dfrac{4.50}{100}=2\left(g\right)\)

     n_MgCl2 = \(\dfrac{m_{MgCl2}}{M_{MgCl2}}=\dfrac{2}{95}=0,02\left(mol\right)\)

d) C⁰/₀ = \(\dfrac{m_{ct}.100}{m_{dd}}\Rightarrow m_{ct}=\dfrac{C.m_{dd}}{100}=\dfrac{15.200}{100}=30\left(g\right)\)

     n_KCl = \(\dfrac{m_{KCl}}{M_{KCl}}=\dfrac{30}{74,5}=0,4\left(mol\right)\)

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