62)

$n_{C_2H_5OH} = 0,5(mol) ; n_{C_2H_5OH\ pư} = 0,5.80\% = 0,4(mol)$
$C_2H_5OH \xrightarrow{t^o,xt} C_2H_4 + H_2O$

$n_{C_2H_4} = n_{C_2H_5OH} = 0,4(mol)$
$m_{C_2H_4} = 0,4.28 = 11,2(gam)$

Câu 63 :

$m_{ancol} = m_{H_2O} + m_{ete} = 5,4 + 19,4 = 24,8(gam)$
$n_{H_2O} = 0,3(mol)$
CTTQ ancol : $C_nH_{2n+1}OH$
$2C_nH_{2n+1}OH \xrightarrow{t^o} C_nH_{2n+1}OC_nH_{2n+1} + H_2O$

$n_{ancol} = 2n_{H_2O} = 0,6(mol)$
$\Rightarrow M_{ancol} = 14n + 18 = \dfrac{24,8}{0,6} = 41,33$
$\Rightarrow n = 1,7$

Vậy hai ancol là $CH_3OH$ và $C_2H_5OH$

\(-6^3=-216\)

\(-6^2=-36\)

\(-6^4=-1296\)

\(-6^5=-7776\)

-6³=-216

-6²= -36

-6⁴= -1296

-6⁵ = -7776

54.113+45.113+113=113.(54+45+1)=113.100=11300

17.(36+62)-17.(62+63)=17.(36+62-62-63)=17.(-27)=-459

11300 và

459

2S = 2 + 2^2 + 2^3 + ...+ 2^64

2S + 1 = 1 + 2 + 2^2 + ... + 2^64

2S - S = 2^64 - 1

Vậy S =  2^64 - 1

\(A=2^2+2^3+...+2^{62}+2^{63}\)

\(2A=2^3+2^4+...+2^{63}+2^{64}\)

\(2A-A=\left(2^3+2^4+...+2^{63}+2^{64}\right)-\left(2^2+2^3+...+2^{62}+2^{63}\right)\)

\(A=2^{64}-2^2\)

mình làm vài câu cho bạn tham khảo,các câu còn lại thì bạn làm tương tự thôi

23.\(\sqrt{14-2\sqrt{33}}=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\left|\sqrt{11}-\sqrt{3}\right|=\sqrt{11}-\sqrt{3}\)

28. \(\sqrt{25-4\sqrt{6}}=\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.1+1^2}=\sqrt{\left(2\sqrt{6}-1\right)^2}\)

\(=\left|2\sqrt{6}-1\right|=2\sqrt{6}-1\)

29.\(\sqrt{14-8\sqrt{3}}=\sqrt{14-2\sqrt{48}}=\sqrt{\left(\sqrt{8}\right)^2-2\sqrt{6}.\sqrt{8}+\left(\sqrt{6}\right)^2}\)

\(=\sqrt{\left(\sqrt{8}-\sqrt{6}\right)^2}=\left|\sqrt{8}-\sqrt{6}\right|=\sqrt{8}-\sqrt{6}\)

nhìn rối mắt quá bạn ơi

+ \(\frac{1}{n\times\left(n+2\right)}=\frac{\left(n+2\right)-n}{n\times\left(n+2\right)}\)

\(=\frac{n+2}{n\times\left(n+2\right)}-\frac{n}{n\times\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)

+ \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)

\(=5-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=5-\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\)

\(=5-\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(=5-\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)

\(=5-\frac{1}{2}+\frac{1}{22}=\frac{50}{11}\)

                              =50/11

Đặt \(A=1+2+2^2+2^3+...+2^{64}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{65}\)

\(\Rightarrow2A-A=2+2^2+2^3+...+2^{65}-\left(1+2+2^2+...+2^{64}\right)\)

\(\Rightarrow A=2+2^2+2^3+...+2^{65}-1-2-2^2-...-2^{64}\)

\(\Rightarrow A=2^{65}-1\)

k có 62 63

Vậy giúp 42 43 đi ạ

đề pẹn ơi

:D?