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a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)
\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-1\right)\)
\(=\dfrac{5}{23}.\dfrac{1}{26}\)
\(=\dfrac{5}{598}\)
b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.1=\dfrac{5}{9}\)
a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)
\(=\dfrac{5}{23}.\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-\dfrac{26}{26}\right)\)
=\(\dfrac{5}{23}.\dfrac{1}{26}\)
\(=\dfrac{5}{598}\)
b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.\left(\dfrac{7}{7}\right)\)
=\(\dfrac{5}{9}.1\)
\(=\dfrac{5}{9}\)
\(a.\)
\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)
\(\Rightarrow-4x=-\dfrac{37}{30}\)
\(\Rightarrow4x=\dfrac{37}{30}\)
\(\Rightarrow x=\dfrac{37}{120}\)
\(b.\)
\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\)
\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)
\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)
\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Rightarrow x=\dfrac{19}{21}\)
\(d.\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)
\(\Rightarrow x=\dfrac{49}{5}\)
\(e.\)
\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)
vậy \(x=\dfrac{37}{120}\)
b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)
c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)
d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)
vậy \(x=\dfrac{49}{5}\)
e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)
th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)
\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)
th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)
\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)
vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)
a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)
= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)
= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)
= \(\dfrac{7}{13}\times\dfrac{7}{12}\)
= \(\dfrac{49}{156}\)
b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)
= 0
Bài 1:
a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)
\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)
c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)
b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{3}{-2}=\dfrac{-9}{6};\dfrac{-1}{-7}=\dfrac{1}{7}\)
Sắp xếp:
\(\dfrac{-1}{-7};\dfrac{0}{8};\dfrac{-7}{6};\dfrac{3}{-2}\)
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\left(4\dfrac{3}{4}+\dfrac{1}{8}+3\dfrac{1}{12}\right)+\left[\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)\right]\)
\(=\left(\dfrac{19}{4}+\dfrac{1}{8}+\dfrac{37}{12}\right)+\left(-4,15\right)\)
\(=\dfrac{191}{24}-4,15\)
\(=\dfrac{457}{120}=3\dfrac{97}{120}\)
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\dfrac{19}{4}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{475}{100}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\dfrac{25}{200}+\left(-\dfrac{256}{200}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\left(-\dfrac{231}{200}\right)+\left(-\dfrac{30}{12}+\dfrac{37}{12}\right)\)
=
Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$
$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$
$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương
$\Rightarrow x+2023=0$
$\Leftrightarrow x=-2023$