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=\(\frac{1.2.3...30.31}{2\left(2.3.4...31\right).64}=\frac{1}{128}\)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
\(30A=\frac{30^{32}+30}{30^{32}+1}=\frac{30^{32}+1+29}{30^{32}+1}=1+\frac{29}{30^{32}+1}\)
\(30B=\frac{30^{33}+30}{30^{33}+1}=\frac{30^{33}+1+29}{30^{33}+1}=1+\frac{29}{30^{33}+1}\)
Vì \(\frac{29}{30^{32}+1}>\frac{29}{30^{33}+1}\) nên \(1+\frac{29}{30^{32}+1}>1+\frac{29}{30^{33}+1}\Rightarrow30A>30B\Rightarrow A>B\)
Vậy \(A>B.\)
Chúc bạn học tốt.
\(S=1+5+5^2+5^4+...+5^{200}\)
\(\Leftrightarrow5^2S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S-S=5^{202}-1\)
\(\Leftrightarrow S=\left(5^{202}-1\right)\div24\)
a) S = 1 + 52 + 54 + ... + 5200
=> 52S = 52.(1 + 52 + 54 + ... + 5200)
=> 25S = 52 + 54 + 56 + ... + 5202
=> 25S - S = (52 + 54 + 56 + ... + 5202) - (1 + 52 + 54 + ... + 5200)
=> 24S = 5202 - 1
=> S = \(\frac{5^{202}-1}{24}\)
a, \(\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}=\frac{2^{13}.\left(2^{17}.5^7+5^{27}\right)}{2^{10}.\left(2^{17}.5^7+5^{27}\right)}=\frac{2^{13}}{2^{10}}=2^3=8\).
b, \(\frac{81.2^2+3^4+20.9^2}{16.3^2+45+2^2.9}=\frac{3^4.2^2+3^4+20.3^4}{16.3^2+3^2.5+2^2.3^2}=\frac{3^4.\left(2^2+1+20\right)}{3^2.\left(16+5+2^2\right)}=\frac{3^4.25}{3^2.25}=\frac{3^4}{3^2}=3^2=9\)
d.
\(-\frac{31}{30}< -1\)
\(-1< -\frac{45}{47}\)
\(\Rightarrow-\frac{31}{30}< -\frac{45}{47}\)
Chúc bạn học tốt
Ta có:
Chọn đáp án D.