Sửa đề thành : S=1+1/5+1/25+.........+1/15625 

5xS=1+1/5+1/25+....+1/3125

5xS-S=(1+1/5+1/25+...+1/3125)-(1/5+1/25+1/125+....+1/15625)

4xS=1-1/15625=15624/15625

1xS=15624/15625:4=15624/15625x1/4=15624/62500=3906/15625

\(#040510\)

\(S=1+\dfrac{1}{5}+\dfrac{1}{25}+...+\dfrac{1}{15625}\)

\(5S=5+1+\dfrac{1}{5}+...+\dfrac{1}{3125}\)

\(5S-S=5-\dfrac{1}{15625}\)

\(S=\dfrac{5.15625-1}{4.15625}\)

\(S=\dfrac{78124}{4.15625}=\dfrac{19531}{15625}\)

a: Xét ΔABC có

M là trung điểm của AB

O là trung điểm của AC

Do đó: MO là đường trung bình của ΔABC

Suy ra: MO=BC/2=3(cm)

a) \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

TheoPT: \(n_{H_2SO_4}=n_{MgO}=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

b) \(m_{ddH_2SO_4}=50.1,2=60\left(g\right)\)

=> \(C\%_{H_2SO_4}=\dfrac{14,7}{60}.100=24,5\%\)

c) \(m_{ddsaupu}=6+60=66\left(g\right)\)

TheoPT: \(n_{MgSO_4}=n_{MgO}=0,15\left(mol\right)\)

=> \(C\%_{MgSO_4}=\dfrac{0,15.120}{66}.100=27,27\%\)

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A

10 giờ 10 phút nha bé iu

A. 10 giờ 10 phút

điền từ hay điền thì vào ạ ????????

Bài III:

1: Ta có: \(\sqrt{x-3}=5\)

\(\Leftrightarrow x-3=25\)

hay x=28

2: Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}-5}=\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-6=\sqrt{x}-5\)

\(\Leftrightarrow2\sqrt{x}=1\)

hay \(x=\dfrac{1}{4}\)