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TH
9 tháng 3 2017
a,
\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)
b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)
c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
17 tháng 9 2017
a) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot2^5\)
\(\Rightarrow2^n\cdot4,5=288\)
\(\Rightarrow2^n=64\)
\(\Rightarrow n=6\)
b) \(2^m-2^n=1984\)
\(\Rightarrow2^n\cdot\left(2^{m-n}-1\right)=2^6\cdot31\)
\(\Rightarrow\left\{{}\begin{matrix}2^n=2^6\\2^{m-n}-1=31\end{matrix}\right.\)
\(\Rightarrow n=6\)
\(\Rightarrow2^{m-n}=32\Rightarrow m-n=5\Rightarrow m=11\)
16 tháng 7 2017
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
LG
16 tháng 7 2017
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
EJ
3 tháng 8 2017
a, \(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{\left(20.5\right)^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=5^5\)
EJ
3 tháng 8 2017
b,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)
TB
17 tháng 12 2017
1) (1+\( \dfrac{2}{3}\)-\(\dfrac{1}{4}\)). 0,8-\(\dfrac{3}{4}\))2
=( \(\dfrac{17}{12}\).0,8-\(\dfrac{3}{4}\))2
=(\(\dfrac{17}{15}\)-\(\dfrac{3}{4}\))2
=( \(\dfrac{23}{60}\))2
= \(\dfrac{529}{3600}\)
31 tháng 7 2022
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
7 tháng 7 2017
C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)
C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
C=1/100-(1-1/100)
C=1/100-99/100
C=-98/100=-49/50
7 tháng 7 2017
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)
25 tháng 10 2018
\(a,-\dfrac{3}{5}.y=\dfrac{21}{10}\)
\(y=\dfrac{21}{10}:\dfrac{-3}{5}=\dfrac{-7}{2}\)
\(b,y:\dfrac{3}{8}=-1\dfrac{31}{33}\)
\(y=-1\dfrac{31}{33}.\dfrac{3}{8}=\dfrac{-8}{11}\)
Vậy \(y=-\dfrac{8}{11}\)
\(c,1\dfrac{2}{5}.y+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\Rightarrow1\dfrac{2}{5}y=-\dfrac{4}{5}-\dfrac{3}{7}=\dfrac{-43}{35}\)
\(\Rightarrow y=\dfrac{-43}{35}:1\dfrac{2}{5}=\dfrac{-43}{49}\)
\(d,-\dfrac{11}{12}.y+0,25=\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{11}{12}.y=\dfrac{5}{6}-0,25=\dfrac{7}{12}\)
\(\Rightarrow y=\dfrac{7}{12}:\dfrac{-11}{12}=\dfrac{-7}{11}\)
QT
Quoc Tran Anh Le
Giáo viên
19 tháng 12 2017
b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{5}{7}\)
QT
Quoc Tran Anh Le
Giáo viên
19 tháng 12 2017
\(\left|x-3\right|=\dfrac{1}{2}\)
\(x-3=\pm\dfrac{1}{2}\)
| x-3 | \(\dfrac{1}{2}\) | \(-\dfrac{1}{2}\) |
| x | \(3\dfrac{1}{2}\) | \(2\dfrac{1}{2}\) |
\(\dfrac{1}{3}\)x 2\(\dfrac{1}{5}\) + \(\dfrac{1}{3}\) x ( -0,8)
= \(\dfrac{1}{3}\) x ( 2\(\dfrac{1}{5}\) - 0,8)
= \(\dfrac{1}{3}\) x ( \(\dfrac{11}{5}\) - \(\dfrac{4}{5}\))
= \(\dfrac{1}{3}\) x \(\dfrac{7}{5}\)
= \(\dfrac{7}{15}\)