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Bài 1:
a) \(0,\left(3\right)+3\dfrac{1}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{19}{45}\)
\(=\dfrac{184}{45}\)
b) \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{1789}{990}\)
Bài 2:
a) \(0,\left(37\right)x=1\)
\(\Leftrightarrow\dfrac{37}{99}.x=1\)
\(\Leftrightarrow x=1:\dfrac{37}{99}\)
\(\Leftrightarrow x=\dfrac{99}{37}\)
b) \(0,\left(26\right)x=1,2\left(31\right)\)
\(\Leftrightarrow\dfrac{26}{99}x=\dfrac{1219}{990}\)
\(\Leftrightarrow x=\dfrac{1219}{990}:\dfrac{26}{99}\)
\(\Leftrightarrow x=\dfrac{1219}{260}\)
Chúc bạn học tốt!
a) \(=\dfrac{\left(-1\right)^4}{3^4}=\dfrac{1}{81}\)
b) \(=\dfrac{\left(-9\right)^3}{4^3}=\dfrac{-729}{64}\)
c) \(=\left(-\dfrac{2}{10}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\)
d) \(=1\)
\(a,=\dfrac{1}{81}\\ b,=\dfrac{729}{64}\\ c,=0,04\\ d,=1\)
a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)
b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)
c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)
d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)
b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)
\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)
hay x=1
a: \(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}\left(1-3-\dfrac{1}{2}\right)\)
\(=\dfrac{-247}{110}+\dfrac{1}{3}\cdot\dfrac{-5}{2}=\dfrac{-508}{165}\)
b: \(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}=\dfrac{139}{90}\)
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
1.
a) \(-\dfrac{4}{9}+\left(-\dfrac{5}{6}\right)-\dfrac{17}{4}=-\dfrac{16}{36}-\dfrac{30}{36}-\dfrac{153}{36}\)
\(=-\dfrac{199}{36}\)
b) \(5\dfrac{1}{2}+\left(-3\right)=5\dfrac{1}{2}-3=\dfrac{11}{2}-\dfrac{6}{2}=\dfrac{5}{2}\)
c) \(4\dfrac{9}{11}+\left(-2\dfrac{1}{11}\right)=\dfrac{53}{11}-\dfrac{23}{11}=\dfrac{30}{11}\)
2.
a) \(4,3-\left(1,2\right)=3,1\)
b) \(0-\left(-0,4\right)=0+0,4=0,4\)
c) \(-\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{3}{3}=-1\)
d) \(-\dfrac{1}{2}-\dfrac{-1}{6}=-\dfrac{1}{2}+\dfrac{1}{6}=-\dfrac{3}{6}+\dfrac{1}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\)
a) = 0,16 - 0,064 . (-3)
= 0,16 + 0,192
= 0,352
b) = (7/4)³ - (7/4)² + 1
= 343/64 - 49/16 + 1
= 147/64 + 1
= 211/64
c) = 8/27 - 4.(-7/4)² - 8/27
= -4.49/16
= -49/4
Lời giải:
a. $=0,16-(-0,064).(-3)=0,16-0,192=-0,032$
b. $=(1\frac{3}{4})^2(1\frac{3}{4}-1)+1=(1\frac{3}{4})^2.\frac{3}{4}+1$
$=\frac{147}{64}+1=\frac{211}{64}$
c.
$=(\frac{2}{3})^3-4(\frac{-7}{4})^2-(\frac{2}{3})^3$
$=-4(\frac{-7}{4})^2=\frac{-49}{4}$
a: Ta có: \(0,\left(3\right)+\dfrac{10}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{4}{9}\)
\(=\dfrac{33}{9}+\dfrac{4}{9}=\dfrac{37}{9}\)
b: Ta có: \(\dfrac{4}{9}+1.2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{440}{990}+\dfrac{1219}{990}-\dfrac{130}{990}\)
\(=\dfrac{139}{90}\)
c: Ta có: \(2,\left(4\right)\cdot\dfrac{3}{11}\)
\(=\dfrac{22}{9}\cdot\dfrac{3}{11}\)
\(=\dfrac{2}{3}\)
d: Ta có: \(-0,\left(3\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
=0