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Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+......................+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.....................+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...................+\dfrac{101}{3^{101}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..............+\dfrac{100}{3^{99}}\right)\)\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+................+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+.............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)-\left(1+\dfrac{1}{3}+..........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Chúc bn học tốt ~
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
\(\left(2^{100}.5+2^{100}.3\right):2^{101}\)
\(=2^{100}.8:2^{101}\)
\(=2^{100}.2^3:2^{101}\)
\(=2^{103}:2^{101}\)
\(=2^2\)
\(=4\)
\(3^5:3^3+2^2.2^3-14\)
\(=3^2+2^6-14\)
\(=9+64-14\)
\(=59\)
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
C = \(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(C=\frac{\left(101+1\right).101:2}{1+1+...+1+1}\)
\(C=\frac{5151}{51}\)
\(C=101\)
b) \(D=\frac{3737.43-4343.37}{2+4+6+...+100}\)
\(D=\frac{37.101.43-43.101.37}{2+4+6+...+100}\)
\(D=\frac{0}{2+4+6+...+100}\)
\(D=0\)
Ta có : A = 3 + 32 + 33 + ..... + 3100
=> 3A = 32 + 33 + ..... + 3101
=> 3A - A = 3101 - 3
=> 2A = 3101 - 3
=> A = \(\frac{3^{101}-3}{2}\)
a) \(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3A-A=3^2+3^3+3^4+...+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}\)
b) Làm tương tự, đáp số là \(B=\frac{3^{106}-3^{100}}{2}\)