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a,
\(\dfrac{5^x}{125}=5^4\\ 5^x:5^3=5^4\\ 5^x=5^4\cdot5^3\\ 5^x=5^7\\ \Rightarrow x=7\)
b,
\(\dfrac{3^x}{3}+3^{x-2}=4\\ 3^{x-1}+3^{x-2}=3^1+3^0\\ \Rightarrow x=2\)
c,
\(\left(x+\dfrac{2006}{2007}\right)^6=0\\ \Rightarrow x+\dfrac{2006}{2007}=0\\ x=0-\dfrac{2006}{2007}\\ x=\dfrac{-2006}{2007}\)
d,
\(\left(x-\dfrac{1}{5}\right)^3=\dfrac{8}{125}\\ \left(x-\dfrac{1}{5}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{1}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}+\dfrac{1}{5}\\ x=\dfrac{3}{5}\)
e,
\(3^x+3^{x-2}=810\\ 3^x\left(1+3^2\right)=810\\ 3^x\cdot10=810\\ 3^x=810:10\\ 3^x=81\\ 3^x=3^4\\ \Rightarrow x=4\)
g,
\(5^{x+2}+5^{x+1}+5^x=19375\\ 5^x\left(5^2+5+1\right)=19375\\ 5^x\cdot31=19375\\ 5^x=19375:31\\ 5^x=625\\ 5^x=5^4\\ \Rightarrow x=4\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
1a, 34.26.15−34.44.15=34.15.(26−44)=320.(−18)=−271034.26.15−34.44.15=34.15.(26−44)=320.(−18)=−2710
b,−√25+2√16=−√27.4=−4√27−25+216=−27.4=−427
2a, 2+x=-5
=>x=-7
b, x−1x+5=67=>7(x−1)=6(x+5)=>7x−7=6x+30=>x=37
1a, \(\frac{3}{4}.26.\frac{1}{5}-\frac{3}{4}.44.\frac{1}{5}=\frac{3}{4}.\frac{1}{5}.\left(26-44\right)=\frac{3}{20}.\left(-18\right)=\frac{-27}{10}\)
b,\(-\sqrt{25+2}\sqrt{16}=-\sqrt{27}.4=-4\sqrt{27}\)
2a, 2+x=-5
=>x=-7
b, \(\frac{x-1}{x+5}=\frac{6}{7}\\ =>7\left(x-1\right)=6\left(x+5\right)\\ =>7x-7=6x+30\\ =>x=37\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a)\(=\frac{-7}{10}+\frac{-1}{5}=\frac{-7}{10}+\frac{-2}{10}=\frac{-9}{10}\)
A = \(\frac{5}{4}.\left(5-\frac{4}{3}\right).\frac{1}{11}\)
A = \(\frac{5}{44}\left(5-\frac{4}{3}\right)\)
A = \(\frac{25}{44}-\frac{5}{33}\)
A = \(\frac{25.3}{4.11.3}-\frac{5.4}{3.11.4}\)
A = \(\frac{75}{132}-\frac{20}{132}\)
A = \(\frac{55}{132}\)
B = \(\frac{3}{4}:\left(-12\right).\left(-\frac{2}{3}\right)\)
B = \(\frac{-1}{16}.\left(-\frac{2}{3}\right)\)
B = \(\frac{1}{24}\)
C = \(\frac{5}{4}:\left(-15\right).\left(-\frac{2}{5}\right)\)
C = \(\frac{-1}{12}.\left(-\frac{2}{5}\right)\)
C = \(\frac{1}{30}\)
D = \(\left(-3\right).\left(\frac{2}{3}-\frac{5}{4}\right):\left(-7\right)\)
D = \(\frac{3}{7}.\left(\frac{8}{12}-\frac{15}{12}\right)\)
D = \(\frac{3}{7}.\left(-\frac{7}{12}\right)\)
D = \(\frac{-3}{12}\)
D = \(\frac{-1}{4}\)
a, (2x - 4)^2 = 36/49
=> 2x - 4 = 6/7 hoặc 2x - 4 = -6/7
=> 2x = 34/7 hoặc x = 22/7
=> x = 34/14 hoặc x = 22/14
b, tương tự a
c, |1 - x| + 0,73 = 3
=> |1 - x| = 2,23
=> 1 - x = 2,23 hoặc 1 - x = -2,23
=> x = -1,23 hoặc x = 3,23
d, tương tự c
a) \(\left(2x-4\right)^2=\frac{36}{49}=\frac{6^2}{7^2}=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow2x-4=\frac{6}{7}\Rightarrow2x=\frac{34}{7}\Rightarrow x=\frac{17}{7}\)
b) \(\left(3x-5\right)^2=\frac{36}{25}=\frac{6^2}{5^2}=\left(\frac{6}{5}\right)^2\)
\(\Rightarrow3x-5=\frac{6}{5}\Rightarrow3x=\frac{31}{5}\Rightarrow x=\frac{31}{15}\)
c)\(\left|1-x\right|+0,73=3\Rightarrow\left|1-x\right|=2,27\)
\(\orbr{\begin{cases}TH1.1-x=2,27\Rightarrow x=-1,27\\TH2.1-x=-2,27\Rightarrow x=3,27\end{cases}}\)
Vậy, x=......
d) \(\left|x+\frac{3}{4}\right|-5=-2\Rightarrow\left|x+\frac{3}{4}\right|=3\)
\(\orbr{\begin{cases}TH1.x+\frac{3}{4}=3\Rightarrow x=\frac{9}{4}\\TH2.x+\frac{3}{4}=-3\Rightarrow x=-3,75\end{cases}}\)
Vậy, x=.......
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