A = \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) + ... + \(\dfrac{1}{7^{100}}\)

7A = 7 + \(\dfrac{1}{7}\) +  \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{7^{100}}\)

7A - A = (7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) +... + \(\dfrac{1}{7^{99}}\)) - (\(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) + ... + \(\dfrac{1}{7^{100}}\))

6A = 7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + ... + \(\dfrac{1}{7^{99}}\) - \(\dfrac{1}{7}\)  - \(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) - ... - \(\dfrac{1}{7^{100}}\)

6A = (\(\dfrac{1}{7}\) - \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^2}\)) + (\(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^3}\)) +...+(\(\dfrac{1}{7^{99}}\) - \(\dfrac{1}{7^{99}}\))+ (7 - \(\dfrac{1}{7^{100}}\))

6A = 0 + 0 + ... + 0 + 7 - \(\dfrac{1}{7^{100}}\)

6A = 7 - \(\dfrac{1}{7^{100}}\)

A = (7 - \(\dfrac{1}{7^{100}}\)) : 6

A = \(\dfrac{7}{6}\) - \(\dfrac{1}{6.7^{100}}\)

   G =   \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\) + ... + \(\dfrac{3}{5^{100}}\)

5³G =  75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) +... + \(\dfrac{3}{5^{99}}\)

125G - G = (75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\) + ... + \(\dfrac{3}{5^{99}}\))  - (\(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\)+\(\dfrac{3}{5^7}\)+...+\(\dfrac{3}{5^{100}}\))

124G =  75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\)+...+ \(\dfrac{3}{5^{99}}\) - \(\dfrac{3}{5}\) - \(\dfrac{3}{5^4}\) - \(\dfrac{3}{5^7}\) - ... - \(\dfrac{3}{5^{100}}\) 

124G = (75 - \(\dfrac{3}{5^{100}}\)) + (\(\dfrac{3}{5}\) - \(\dfrac{3}{5}\)) +(\(\dfrac{3}{5^4}\) - \(\dfrac{3}{5^4}\)) +...+ (\(\dfrac{3}{5^{99}}\) - \(\dfrac{3}{5^{99}}\)) 

124G = 75 - \(\dfrac{3}{5^{100}}\) + 0 + 0 + ... + 0

124G =  75 - \(\dfrac{3}{5^{100}}\)

   G = (75 - \(\dfrac{3}{5^{100}}\)): 124

   G = \(\dfrac{75}{124}\) - \(\dfrac{3}{124.5^{100}}\)

Thể tích bể nước: \(3\left(3-1,8\right)\)\(.1,5\) \(=3.1,2.1,5\)\(=5,4m^3\)

Đổi \(5,4m^3\) \(=5400dm^3\)

\(=>\) Bể chứa được \(5400l\)

:)

:)

Ta có: \(\dfrac{1}{500}\)\(m^3\) \(=\) \(\dfrac{2}{1000}\)\(m^3\) \(=2dm^3\)

\(=>2dm^3=2.1000=2000cm^3\)

Cảm ơn 😀:))

\(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\\ a,PTHH:CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b,n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,5\left(mol\right)\\ V_{CO_2\left(25^oC,1bar\right)}=0,5.24,79=12,395\left(l\right)\\ c,n_{HCl}=2.0,5=1\left(mol\right)\\ V_{ddHCl}=\dfrac{1}{0,2}=5\left(l\right)\\ d,m_{CaCl_2}=111.0,5=55,5\left(g\right)\)

em thêu vải ngắn hơn bà.

bà em thêu vải dài hơn em

Lời giải:

Chiều cao tam giác là: $30\times 2:3=20$ (cm)

Độ dài đáy tam giác:

$120\times 2:20=12$ (cm)