giải cái gì bạn

Bài 2:

ΔABC vuông tại A

=>\(cosC=\sin B=0,2\)

TA có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-0,2^2=1-0,04=0,96\)

=>\(\sin C=\sqrt{0,96}=\sqrt{\frac{24}{25}}=\frac{2\sqrt6}{5}\)

tan C=\(\frac{\sin C}{cosC}=\frac{2\sqrt6}{5}:\frac15=2\sqrt6\)

cot C=\(\frac{1}{2\sqrt6}=\frac{\sqrt6}{12}\)

Câu 3:

Ta có: \(\tan B=\frac{1}{\cot B}\)

=>\(\frac{1}{\cot B}=5\)

=>\(\cot B=\frac15\)

Ta có: \(1+\tan^2B=\frac{1}{cos^2B}\)

=>\(\frac{1}{cos^2B}=1+5^2=26\)

=>\(cos^2B=\frac{1}{26}\)

=>\(cosB=\frac{1}{\sqrt{26}}\)

Ta có: \(\sin^2B+cos^2B=1\)

=>\(\sin^2B=1-\frac{1}{26}=\frac{25}{26}\)

=>\(\sin B=\frac{5}{\sqrt{26}}\)

13 B

14 B

15 C

cái đề đâu bn

1 Mary asked me who I talked to when i had problems

2 Hoa said she would help her mum cook dinner that night

3 jack advised me to tell my teacher what had happened

4 Nam said his best friend hadn't called him for one week

5 Lucia's mother asked her if she was at the sports center then

6 Tom asked mark what time he had come home the night before

7 Mrs Brown told me not to go to the park when it gets dark

8 Mrs QUang told Trung they had spoken to his parents the day before

9 Minh asked Phuong if he could meet her at 4.30 the day after afternoom

10 Nga said she was staying with her aunt and uncle in the suburbs

1. Mary asked me who I talked to when I had problems.

2. Hoa said that she would help her mum cook dinner that night.

3. Jack advised me to tell my teacher what had happened.

4. Nam said that his best friend hadn't called him for one week.

5. Lucia's mother asked her if she were at the sports centre then.

6. Tom asked Mark what time he had come home the previous night.

7. Mrs. Brown told me not to go to the park when it got dark.

8. Mr. Quang said to Trung that they had spoken to his parents the day before.

9. Minh asked Phuong if he could met her at 4.30 that next afternoon.

10. Nga said that she was staying with her aunt and uncle in the suburbs.

Câu 3:

a) 

CTPT xủa X là C_nH_2n+2O

\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\Rightarrow n_{C_nH_{2n+2}O}=\dfrac{0,4}{n}\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{\dfrac{0,4}{n}.\left(2n+2\right)}{2}=\dfrac{0,4}{n}\left(n+1\right)\left(mol\right)\)

Mà \(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)

=> n = 4

=> CTPT: C₄H₁₀O

b) \(n_{C_4H_{10}O}=\dfrac{0,4}{4}=0,1\left(mol\right)\)

=> m = 0,1.74 = 7,4 (g)

c)

(1) \(CH_3-CH_2-CH_2-CH_2OH\)

(2) \(CH_3-CH_2-CH\left(OH\right)-CH_3\)

(3) \(CH_3-C\left(CH_3\right)\left(OH\right)-CH_3\)

(4) \(CH_3-CH\left(CH_3\right)-CH_2OH\)

(5) \(CH_3-CH_2-CH_2-O-CH_3\)

(6) \(CH_3-CH\left(CH_3\right)-O-CH_3\)

(7) \(CH_3-CH_2-O-CH_2-CH_3\)

d)

X là \(CH_3-C\left(CH_3\right)\left(OH\right)-CH_3\) (2-metylpropan-2-ol)

1) \(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}:\dfrac{\sqrt{x}-1}{5}\)

        \(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{5}{\sqrt{x}-1}\) \(=\dfrac{5}{x+\sqrt{x}+1}\)

2) Ta thấy \(x+\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}+1\right)+1>1\forall x\)

\(\Rightarrow A< 5\)

1/ \(n_S=\dfrac{6,4}{32}=0,2;n_{H_2SO_4}=\dfrac{14.70\%}{98}=0,1\)

Bảo toàn nguyên tố S : \(n_S=n_{H_2SO_4\left(lt\right)}=0,2\)

Mà thực tế chỉ thu được 0,1

=> \(H=\dfrac{0,1}{0,2}.100=50\%\)

2/ \(n_{N_2}=0,2\left(mol\right);n_{H_2}=0,3\left(mol\right);n_{NH_3}=0,15\left(mol\right)\)

PTHH: \(N_2+3H_2\rightarrow2NH_3\)

Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,3}{3}\)=> Sau phản ứng N2 dư, tính theo số mol H2

=> n NH3(lt)= \(\dfrac{0,3.2}{3}=0,2\left(mol\right)\)

Mà thực tế chỉ thu được 0,15 mol 

=> \(H=\dfrac{0,15}{0,2}.100=75\%\)