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a) \(A=\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(2^2.5\right)^4}{5^{2^5}.\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{\left(5^{10}:5^8\right).\left(2^{10}:2^8\right)}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)\(=\frac{2^3+2^3}{1}=\frac{8+8}{1}=16\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\right)\)
\(\Rightarrow C=1-\frac{1}{2^{100}}\)
d) \(D=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{100}}\)
\(\Rightarrow5D=5+1+\frac{1}{5^2}+\frac{1}{5^3}+...........+\frac{1}{5^{101}}\)
\(\Rightarrow5D-D=\left(5+1+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{101}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..........+\frac{1}{5^{100}}\right)\)
\(\Rightarrow4D=5-\frac{1}{5^{101}}\)
\(\Rightarrow D=\frac{5-\frac{1}{5^{101}}}{4}\)
a) \(A=\frac{5^4x20^4}{25^5x4^5}=\frac{5^4x\left(2^2x5\right)^4}{\left(5^2\right)^5x\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2x2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}=\frac{2^{13}.5^7.\left(2^{17}+5^{20}\right)}{2^{10}.5^7.\left(2^{17}+5^{20}\right)}=2^3=8\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=1-\frac{1}{2^{100}}\)
\(C=1-\frac{1}{2^{100}}\)
phần d bn lm tương tự như phần c nha!
\(a,8^7-2^{18}=2^{21}-2^{18}=2^{17}\left(2^4-2\right)=14.2^7⋮14\)
\(b,10^6-5^7=2^6.5^6-5^7=5^6\left(2^6-5\right)=59.5^6⋮59\)
c ko nói chia hết cho số nào
\(d,3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^{n+1}\left(3^2+3\right)+2^{n+1}\left(2^2+2\right)\)
\(=3^{n+1}.12+2^{n+1}.6\)
\(=6.\left(3^{n+1}.2+2^{n+1}\right)⋮6\)
\(S=1+2\cdot3+3\cdot3^2+4\cdot3^3+...+101\cdot3^{100}=\left(1+3+3^2+...+3^{100}\right)+\left(3+3^2+...+3^{100}\right)+...+3^{100}\)\(S=\left(1+...+3^{100}\right)+3\left(1+...+3^{99}\right)+3^2\left(1+...+3^{98}\right)+...+3^{100}\)
\(S=1\cdot A_{100}+3\cdot A_{99}+3^2\cdot A_{98}+...+3^{100}\)
\(A_i=1+3+3^2+...+3^i=\frac{3^{i+1}-1}{2}\)
\(S=\frac{3^{101}-1}{2}+\frac{3\left(3^{100}-1\right)}{2}+\frac{3^2\left(3^{99}-1\right)}{2}+...+\frac{3^{100}\left(3-1\right)}{2}\)
\(2S=3^{101}\cdot101-\left(1+2+3+...+3^{100}\right)=101\cdot3^{101}-A_{100}=101\cdot3^{101}-\frac{3^{101}-1}{2}\)
\(2S=\frac{201\cdot3^{101}+1}{2}\Leftrightarrow S=\frac{201\cdot3^{101}+1}{4}\)
2\(^3\)+3.(\(\frac{1}{9}\))\(^6\)-2\(^{-2}\).4+[(-2)\(^2\):\(\frac{1}{2}\)].8
Gấp mk cần gấp!
HELP ME!
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
a: \(\left(x-6\right)^2=\left(x-6\right)^3\)
\(\Leftrightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Leftrightarrow\left(x-6\right)^2\left(x-7\right)=0\)
hay \(x\in\left\{6;7\right\}\)
b: \(3+2^{x-1}=24-4^2-\left(2^2-1\right)\)
\(\Leftrightarrow2^{x-1}=24-16-3-3=8-6=2\)
=>x-1=2
hay x=3
a, 2S=2+\(2^2\)+\(2^3+...+2^{201}\)
2S-S=S=\(2^{201}-1\)
câu b thì nhân 3 rồi làm tương tự nha. tiickk mik vs