a: A=1/2+1/4+...+1/1024

=>2A=1+1/2+...+1/512

=>2A-A=1-1/1024=1023/1024

=>A=1023/1024

b: \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=1-1/100=99/100

c: \(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)

=1/2-1/11

=9/22

d: Để (d1) vuông góc với y=(k-1)x+4 thì \(\left(k-1\right)\left(k-3\right)=-1\)

\(\Leftrightarrow k=2\)

a: \(4-\sqrt{3-2x}=0\)

\(\Leftrightarrow3-2x=16\)

hay \(x=-\dfrac{13}{2}\)

\(a) Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{H_2} = n_{Mg} = \dfrac{9,6}{24} = 0,4(mol)\\ V_{H_2} = 0,4.22,4 = 8,96(lít)\\ b) n_{(CH_3COO)_2Mg} = n_{Mg} = 0,4(mol)\\ m_{Muối} = 0,4.142 = 56,8(gam)\\ c) n_{CH_3COOH} = 2n_{Mg} = 0,8(mol)\\ m_{dd\ CH_3COOH} = \dfrac{0,8.60}{6\%} = 800(gam)\\ d) C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ n_{C_2H_5OH} = n_{CH_3COOH} = 0,8(mol)\\ m_{C_2H_5OH} = 0,8.46 = 36,8(gam)\)

mMg=9,6/24=0,4 mol

Mg + 2CH3COOH --> (CH3COO)2Mg+H2

0,4         0,8                   0,4                   0,4                     mol

=>VH2=0,4*22,4=8,96 lít

=>m(CH3COO)2Mg=0,4*142=56,8 g

mddCH3COOH=0,8*60/6%=800 g

C2H5OH +O2 -men giấm->CH3COOH + H2O

 0,8                                          0,8       mol

mC2H5OH=0,8*46=36,8G

1. bought

2. character

3. went

4. found

5. inside

1. bought

2. character

3. went

4. found

5. inside