Tính tổng \(S=1+0,9+\left(0,9\right)^2+\left(0,9\right)^3+....+\left(0,9\right)^{n-1}+...\)
tính giới hạn sau
a, lim\(\frac{1}{\sqrt{n+2}-\sqrt{n+1}}\)
b, lim\(\frac{8^{2n+3}-3^{3n+2}}{4^{3n+4}+5^{2n+3}}\)
tính tổng CSN: \(1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8},...,\left(-\frac{1}{2}\right)^{n-1},...\)
tính tổng S= \(1+0,9+\left(0,9\right)^2+\left(0,9\right)^3+...+\left(0,9\right)^{n-1}+...\)
\(A=lim\frac{\sqrt{n+2}+\sqrt{n+1}}{1}=lim\left[n\left(\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}\right)\right]=+\infty.2=+\infty\)
\(B=lim\frac{8^3.64^n-9.27^n}{4^4.64^n+5^3.25^n}=\frac{8^3-9.\left(\frac{27}{64}\right)^n}{4^4+5^3\left(\frac{25}{64}\right)^n}=\frac{8^3}{4^4}=2\)
\(1;-\frac{1}{2};\frac{1}{4}...\) là dãy cấp số nhân lùi vô hạn có \(u_1=1\) và \(q=-\frac{1}{2}\)
Do \(\left|q\right|< 1\) nên theo công thức tổng cấp số nhân:
\(S_n=\frac{u_1}{1-q}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\)
câu tính tổng S mk làm đc oy nhé k cần lm câu đó nữa đâu
Tính
\(\frac{\left(-5\right).\left(-0,9\right)^{2^{ }}}{\left(1\frac{1}{2}\right)^{4^{ }}.\left(-3\frac{1}{3}\right)^{3^{ }}.\left(-1\right)^{2015}}\)
\(=\dfrac{\left(-5\right)\cdot0.81}{\left(\dfrac{5}{2}\right)^4\cdot\left(-\dfrac{10}{3}\right)^3\cdot\left(-1\right)}=\dfrac{-5\cdot0.81}{\dfrac{5^4\cdot10^3}{2^4\cdot3^3}}\)
\(=-4.05:\dfrac{5^7}{3^3\cdot2}=\dfrac{-81}{20}\cdot\dfrac{3^3\cdot2}{5^7}=\dfrac{-3^7\cdot2}{2^2\cdot5^8}=\dfrac{-3^7}{2\cdot5^8}\)
\(\frac{\left(-5\right)^3.\left(0,9\right)^2}{\left(\frac{3}{2}\right)^4\left(-\frac{10}{3}\right)^3.\left(-1\right)^7}\)
C = \(\frac{\left(-5\right)^3.\left(-0,9\right)^2}{\left(1\frac{1}{2}\right)^4.\left(-3\frac{1}{3}\right)^3.\left(-1\right)^7}\)
Lm giúp mk nhé, thk các bn nhìu <33
\(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)tính gtbt
\(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}=\frac{\left(3.0,3\right)^5}{\left(0,3\right)^6}=\frac{3^5.\left(0,3\right)^5}{\left(0,3\right)^6}=\frac{3^5}{0,3}=\frac{243}{0,3}=810\)
BÀI 2 TÍNH
A = \(\dfrac{20^5.5^{10}}{100^5}\)
B = \(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
GIÚP MÌNH VỚI MN ƠIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
\(A=\dfrac{\left(20.5\right)^5.5^5}{100^5}=\dfrac{100^5.3125}{100^5}=3125\)
\(B=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^5.0,3}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^5.0,3}=\dfrac{3^5}{0,3}=810\)
\(A=\dfrac{2^{10}\cdot5^5\cdot5^{10}}{2^{10}\cdot5^{10}}=5^5=3125\\ B=\left(\dfrac{0,9}{0,3}\right)^5\cdot\dfrac{1}{0,3}=\dfrac{3^5}{0,3}=\dfrac{243}{0,3}=810\)
Tính:
\(\left\{\left[\left(6,2:0,31-\dfrac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}\right]\)
b) \(0,4\left(3\right)+0,6\left(2\right).2\dfrac{1}{2}.\left[\left(\dfrac{1}{2}+\dfrac{1}{3}\right):0,5\left(8\right)\right]:\dfrac{50}{53}\)
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
Tính:
a)\(\left\{\left[\left(6,2:0,31-\frac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\frac{4}{11}:0,1\right).\frac{1}{33}\right]\)
b)\(0,4\left(3\right)+0,6\left(2\right)-2\frac{1}{2}.\left[\left(\frac{1}{2}+\frac{1}{3}:0,5\left(8\right)\right)\right]:\frac{50}{53}\)
c)\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
Tính:
a)\(\frac{{ - 1}}{6} + 0,75\);
b)\(3\frac{1}{{10}} - \frac{3}{8}\);
c)\(0,1 + \frac{{ - 9}}{{17}} - \left( { - 0,9} \right)\).
a)\(\frac{{ - 1}}{6} + 0,75 = \frac{{ - 1}}{6} + \frac{3}{4} = \frac{{ - 2}}{{12}} + \frac{9}{{12}} = \frac{7}{{12}}\);
b)\(3\frac{1}{{10}} - \frac{3}{8} = \frac{{31}}{{10}} - \frac{3}{8} = \frac{{124}}{{40}} - \frac{{15}}{{40}} = \frac{{109}}{{40}}\);
c)
\(\begin{array}{l}0,1 + \frac{{ - 9}}{{17}} - \left( { - 0,9} \right) = \frac{1}{{10}} + \frac{{ - 9}}{{17}} + \frac{9}{{10}}\\ = (\frac{1}{{10}} + \frac{9}{{10}}) + \frac{{ - 9}}{{17}} = 1 + \frac{{ - 9}}{{17}} =\frac{{ 17}}{{17}}+\frac{{ - 9}}{{17}}= \frac{8}{{17}}\end{array}\)