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\(a.\)
\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)
\(\Rightarrow-4x=-\dfrac{37}{30}\)
\(\Rightarrow4x=\dfrac{37}{30}\)
\(\Rightarrow x=\dfrac{37}{120}\)
\(b.\)
\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\)
\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)
\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)
\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Rightarrow x=\dfrac{19}{21}\)
\(d.\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)
\(\Rightarrow x=\dfrac{49}{5}\)
\(e.\)
\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
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 \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
1) \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}\)
\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\\ =\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\\ =\dfrac{12}{7}\cdot\dfrac{35}{8}\\ =\dfrac{15}{2}\)
2) \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}\cdot5\\ =\dfrac{1}{3}\cdot1-\dfrac{2}{3}\\ =\dfrac{1}{3}-\dfrac{2}{3}\\ =-\dfrac{1}{3}\)
3) \(\dfrac{4}{9}\cdot19\dfrac{1}{3}-\dfrac{4}{9}\cdot39\dfrac{1}{3}\)
\(=\dfrac{4}{9}\left(19\dfrac{1}{3}-39\dfrac{1}{3}\right)\\ =\dfrac{4}{9}\cdot\left(\dfrac{58}{3}-\dfrac{118}{3}\right)\\ =\dfrac{4}{9}\cdot\left(-20\right)\\ =-\dfrac{80}{9}\)
a) \(\left(\dfrac{-3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-9}{4}\right):\dfrac{3}{7}\)
\(=\left(\dfrac{-3}{4}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{-9}{4}\right):\dfrac{3}{7}\)
\(=-2:\dfrac{3}{7}=\dfrac{-14}{3}\)
\(\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{8}:\dfrac{1}{6}+\dfrac{7}{8}:\dfrac{-7}{18}\)
\(=\dfrac{7}{8}:\left(\dfrac{1}{6}+\dfrac{-7}{18}\right)=\dfrac{7}{8}:\dfrac{-2}{9}=\dfrac{63}{-16}\)
Các bạn ơi,mình ghi thiếu,còn 3 câu nữa nha!!!~~nya
e)| \(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\) |-(-22).\(\dfrac{1}{3}\)(0,75-\(\dfrac{1}{7}\))=\(\dfrac{-5}{13}\):2\(\dfrac{9}{13}\)-0,5.(\(\dfrac{-2}{3}\))
f)| 5x+21 | = | 2x -63 |
g) -45 - |-3x-96 | - 54=-207
Làm ơn giúp mình với ạ!Mình đang cần gấp lắm trong ngày hôm nay ạ!!!Mình xin cảm ơn các bạn nhiều nhiều lắm luôn đó!!!Thank you very much!!!(^-^)
a, (\(\dfrac{2}{9}\)(6x - \(\dfrac{3}{4}\)) - 3(\(\dfrac{1}{4}x-\dfrac{1}{5}\)) = \(\dfrac{-8}{15}\)
<=> (\(\dfrac{4}{3}x-\dfrac{1}{6}\)) - (\(\dfrac{3}{4}x-\dfrac{3}{5}\)) = \(\dfrac{-8}{15}\)
<=> \(\dfrac{4}{3}x-\dfrac{1}{6}-\dfrac{3}{4}x+\dfrac{3}{5}=\dfrac{-8}{15}\)
<=> \(\dfrac{7}{12}x+\dfrac{13}{30}=\dfrac{-8}{15}\)
<=> \(\dfrac{7}{12}x=\dfrac{-8}{15}-\dfrac{13}{30}\)
<=> \(\dfrac{7}{12}x=-\dfrac{29}{30}\)
<=> x = \(-\dfrac{58}{35}\)
@Nguyễn Gia Hân
1. Tìm \(x\):
a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=1\)
b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)
\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)
\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)
\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)
\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)
\(x=\dfrac{-17}{8}\)
c) \(2016^3.2016^x=2016^8\)
\(2016^x=2016^8:2016^3\)
\(2016^x=2016^{8-3}\)
\(2016^x=2016^5\)
\(\Rightarrow x=5\)
d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)
\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)
\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)
\(x+\dfrac{3}{4}=\dfrac{35}{4}\)
\(x=\dfrac{35}{4}-\dfrac{3}{4}\)
\(x=\dfrac{32}{4}=8\)
e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)
\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)
\(2,8.x-2^5=6\)
\(2,8.x=6+32\)
\(2,8.x=38\)
\(x=38:2,8\)
\(x=\dfrac{95}{7}\)
f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)
\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}:\dfrac{4}{7}\)
\(x=\dfrac{28}{15}\)
g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)
\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}\)
\(\Rightarrow3x=-6\)
\(x=\left(-6\right):3\)
\(x=-2\)
2. Thực hiện phép tính:
a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)
\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)
\(=\dfrac{7}{18}+\dfrac{9}{5}\)
\(=\dfrac{197}{90}\)
b) \(\dfrac{7.5^2-7^2}{7.24+21}\)
\(=\dfrac{7.25-7.7}{7.24+7.3}\)
\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)
\(=\dfrac{7.18}{7.27}\)
\(=\dfrac{2}{3}\)
c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
Các bn ơi nhanh lên nha
\(\left(x+1,5-\dfrac{3}{7}\right):\dfrac{1}{2}=1\dfrac{1}{7}\)
=> \(\left(x+\dfrac{3}{2}-\dfrac{3}{7}\right).2=\dfrac{8}{7}\)
=> \(x+\dfrac{3}{2}-\dfrac{3}{7}=\dfrac{8}{7}.\dfrac{1}{2}\)
=> \(x+\dfrac{21}{14}-\dfrac{6}{14}=\dfrac{8}{14}\)
=> x=\(\dfrac{8}{14}+\dfrac{6}{14}-\dfrac{21}{14}\)
=> x=\(\dfrac{-7}{14}=-\dfrac{1}{2}\)
vậy x=-\(\dfrac{1}{2}\)