Mình sửa lại đề bài là AB cắt CD tại T chứ không phải là AD cắt BC đâu. 

\(CH_3COOCH_3+NaOH\rightarrow CH_3COONa+CH_3OH\\ n_{NaOH}=n_{CH_3COOCH_3}=\dfrac{7,4}{74}=0,1\left(mol\right)\\ Vậy:a=m_{ddNaOH}=\dfrac{0,1.40.100}{4}=100\left(g\right)\)

\(a,cos\alpha=\dfrac{5}{13}\)

\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{5}{13}\right)^2}\Leftrightarrow tan^2\alpha=\dfrac{144}{25}\Leftrightarrow tan\alpha=\dfrac{12}{5}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{12}{5}=\dfrac{5}{12}\)

\(b,sin\alpha=\dfrac{7}{12}\)

\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(\dfrac{7}{12}\right)^2}=\dfrac{\sqrt{95}}{12}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{\sqrt{95}}{12}\right)^2}\Leftrightarrow tan\alpha=\dfrac{49}{95}\)

\(cot\alpha=1:\dfrac{49}{95}=\dfrac{95}{49}\)

\(c,tan\alpha=\dfrac{15}{4}\)

\(cot\alpha=1:\dfrac{15}{4}=\dfrac{4}{15}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(\dfrac{15}{4}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\sqrt{\dfrac{16}{241}}\)

\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\sqrt{\dfrac{16}{241}}\right)^2}\approx0,97\)

\(d,cot\alpha=-\dfrac{1}{\sqrt{3}}\\ tan\alpha=1:\left(-\dfrac{1}{\sqrt{3}}\right)=-\sqrt{3}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(-\sqrt{3}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\dfrac{1}{2}\)

\(sin\alpha=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)

1e+84937

Ta có x_n luôn dương

Ta có \(2x_n+1=\) \(2\times\dfrac{\left(2+cos\alpha\right)x_n+cos^2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}+1=\)

\(=\dfrac{6x_n+2cos^2\alpha+2-cos2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}\)

\(=\dfrac{6x_n+2cos^2\alpha+2sin^2a+1}{\left(2x_n+1\right)\left(1-cos2\alpha\right)+1}\)

\(=\dfrac{3\left(2x_n+1\right)}{2\sin^2\alpha\left(2x_n+1\right)+1}\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}=\dfrac{2\sin^2\alpha\left(2x_n+1\right)+1}{3\left(2x_n+1\right)}\)

\(=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right)\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right)\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{2x_1+1}-\sin^2\alpha\right)\)

\(=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{3}-\sin^2\alpha\right)\)

\(\Rightarrow y_n=\sum\limits^{n-1}_{i=0}\left(\dfrac{1}{3}\right)^i\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)

\(=\dfrac{1-\left(\dfrac{1}{3}\right)^n}{1-\dfrac{1}{3}}\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)

Ta có \(\overrightarrow{BH}=\overrightarrow{BC}+\overrightarrow{CH}=-\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{CA}\)

\(\overrightarrow{MH}=\dfrac{1}{2}\overrightarrow{BA}=\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CA}\right)=\dfrac{1}{2}\overrightarrow{CA}-\dfrac{1}{2}\overrightarrow{CB}\)

\(x_1=a>2;x_{n+1}=x_n^2-2,\forall n=1,2,...\)

mà \(n\rightarrow+\infty\)

\(\Rightarrow a\rightarrow+\infty\Rightarrow x_n\rightarrow+\infty\)

\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\dfrac{1}{x_n}=0\) \(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_nx_{n+1}}\right)=0\)

\(\)\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_1}+\dfrac{1}{x_1x_2}+\dfrac{1}{x_1x_2x_3}+...+\dfrac{1}{x_1x_2...x_n}\right)=0\)

...

41.             C. debatable

42.             B. continual

43.             A. limited

44.             D. inclined

45.             B. economical

46.             B. renowned

47.             B. confined

48.             A. likelihood

49.             A. Favourable

50.             C. minimal

41.             C. debatable

42.             B. continual

43.             A. limited

44.             D. inclined

45.             B. economical

46.             B. renowned

47.             B. confined

48.             A. likelihood

49.             A. Favourable

50.             C. minimal

Đèn cháy sáng chứng tỏ dung dịch dẫn điện. Trong dung dịch có các ion.

Rất dễ em đừng lo lắng. Khi tăng áp suất cân bằng chuyển dịch theo chiều giảm số mol phân tử khí

Hiện tại sách là chương trình mới những vẫn không có PowerPoint bạn nha. Nên hiện tại chưa biết được khi nào sẽ có PowerPoint, nhưng ở các lớp dưới thì cũng đã học rồi á bạn.

Theo chương trình sgk mới thì lớp 7, lớp 8 có học PowerPoint. Lớp 11 thì chưa có em nhé!