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3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50
a; \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + ... + \(\dfrac{5}{132}\)
= 5.(\(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ..+ \(\dfrac{1}{132}\))
= 5.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{11.12}\))
= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))
= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))
= 5.(\(\dfrac{6}{12}\) - \(\dfrac{1}{12}\))
= 5.\(\dfrac{5}{12}\)
= \(\dfrac{25}{12}\)
Bài 1:
a) x^3 = -27 = (-3)^3
=> x = -3
b) (2x-1)^3 = 8^3
=> 2x-1=8
2x = 9
x = 9/2
c) (2x-3)^2 = 9 = 3^2 = (-3)^2
=> 2x -3 = 3 => 2x = 6 => x = 3
2x - 3 = -3 => 2x = 0 => x = 0
KL:...
Bài 2:
a) \(\frac{2563.25^2}{5^{10}}=\frac{2563.5^4}{5^{10}}=\frac{2563}{5^6}\)
\(\frac{2^8.9^2}{6^4.8^2}=\frac{2^8.3^4}{2^{10}.3^4}=\frac{1}{2^2}=\frac{1}{4}\)
Bài 3:
ta có: (-5) < (-3)
=>(-5)^10 < (-3)^10
(-3/4)63x-1=(3/4)^3
3x-1=3+1
3x=3=1
x=4;3
x=4/3
Vậy x=4/3
a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}<\frac{1}{11^{21}}\) Vì 2733 > 1133 > 1121
a) \(\left(x-3\right)^3=27\)
\(\Leftrightarrow x-3=3\)
\(\Rightarrow x=6\)
b) \(x^3=x^6\)
\(\Leftrightarrow x^6-x^3=0\)
\(\Leftrightarrow x^3\left(x^3-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) \(\left(2x+2\right)^5=\left(2x+2\right)^6\)
\(\Leftrightarrow\left(2x+2\right)^5\left(2x+2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x+2\right)^5=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-\frac{1}{2}\end{cases}}\)
(4/5-1/6) x = -3/4.8/27
⇔ 19/30 x = -2/9
⇔ x =-20/57