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a) Đặt \(C=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)
\(\Rightarrow5C=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)
\(\Rightarrow5C-C=1-\dfrac{1}{5^{100}}\Rightarrow4C=1-\dfrac{1}{5^{100}}\Rightarrow C=\dfrac{1-\dfrac{1}{5^{100}}}{4}\)
\(\Rightarrow A=8.5^{100}.\dfrac{1-\dfrac{1}{5^{100}}}{4}+1=2.\left(5^{100}-1\right)+1=2.5^{100}-2+1=2.5^{100}-1\)
b)\(B=\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
\(B=4.\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)\)
Đặt \(\left(\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\right)=D\)
\(\Rightarrow3D=1-\dfrac{1}{3}+...-\dfrac{1}{3^{99}}\)
\(\Rightarrow3D+D=1-\dfrac{1}{3^{100}}\)
\(\Rightarrow D=\dfrac{1-\dfrac{1}{3^{100}}}{4}\)

\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{99}{2^{100}}+\frac{100}{2^{101}}\)
\(A-A2=\frac{1}{2}A=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)-\frac{100}{2^{101}}\)
\(=\frac{\left[1-\left(\frac{1}{2}\right)^{10}\right]}{\left(1-\frac{1}{2}\right)}-\frac{100}{2^{101}}\)
\(=\frac{\left(2^{101-1}\right)}{2^{100}}-\frac{100}{2^{101}}\)
\(\Rightarrow A=\frac{\left(2^{101-1}\right)}{2^{99}}-\frac{100}{2^{100}}\)

Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)

\(A=1+2^1+2^2+......+2^{2006}\)
\(2A=2.\left(1+2^1+2^2+......+2^{2006}\right)\)
\(2A=2+2^2+2^3+........+2^{2007}\)
\(2A-A=\left(2+2^2+2^3+....+2^{2007}\right)-\left(1+2+2^2+...+2^{2006}\right)\)
\(A=2^{2007}-1\)
\(B=1+3+3^2+.....+3^{100}\)
\(3B=3.\left(1+3+3^2+......+3^{100}\right)\)
\(3B=3+3^2+3^3+.....+3^{101}\)
\(3B-B=\left(3+3^2+3^3+....+3^{101}\right)-\left(1+3+3^2+....+3^{100}\right)\)
\(B=3^{101}-1\)
Các phần còn lại bạn làm tương tự như trên nha

Đặt là a, b nhá
\(a)\) \(7^{x-1}-2.7^{100}=5.7^{100}\)
\(\Leftrightarrow\)\(7^{x-1}=5.7^{100}+2.7^{100}\)
\(\Leftrightarrow\)\(7^{x-1}=7^{100}\left(5+2\right)\)
\(\Leftrightarrow\)\(7^{x-1}=7^{100}.7\)
\(\Leftrightarrow\)\(7^{x-1}=7^{101}\)
\(\Leftrightarrow\)\(x-1=101\)
\(\Leftrightarrow\)\(x=101+1\)
\(\Leftrightarrow\)\(x=102\)
Vậy \(x=102\)
\(b)\) \(5^{x-4}=25\)
\(\Leftrightarrow\)\(5^{x-4}=5^2\)
\(\Leftrightarrow\)\(x-4=2\)
\(\Leftrightarrow\)\(x=2+4\)
\(\Leftrightarrow\)\(x=6\)
Vậy \(x=6\)
Chúc bạn học tốt ~
\(7^{x-1}-2.7^{100}=5.7^{100}\)
\(\Rightarrow7^{x-1}=5.7^{100}+2.7^{100}\)
\(\Rightarrow7^{x-1}=7.7^{100}\)
\(\Rightarrow7^{x-1}=49^{100}\)
\(\Rightarrow7^{x-1}=7^{2^{100}}\)
\(\Rightarrow7^{x-1}=7^{200}\)
\(\Rightarrow x=201\)
Vậy x = 201
\(5^{x-4}=25\)
\(\Rightarrow5^{x-4}=5^2\)
\(\Rightarrow x=6\)
Vậy x = 6
\(\left(2^{100}.5+2^{100}.3\right):2^{101}\)
\(=2^{100}.8:2^{101}\)
\(=2^{100}.2^3:2^{101}\)
\(=2^{103}:2^{101}\)
\(=2^2\)
\(=4\)
\(3^5:3^3+2^2.2^3-14\)
\(=3^2+2^6-14\)
\(=9+64-14\)
\(=59\)
là sao v bn?
=\(\frac{2}{2^2.3^2.4^2.5^2\cdot\ldots\cdot100^2}\)