Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(0,5\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\left(-\dfrac{4}{5}\right)\)
\(\Rightarrow\dfrac{1}{2}\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{1}{2}=x-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{2}{3}=x-\dfrac{4}{5}\)
\(\Leftrightarrow15x-20=30x-24\)
\(\Leftrightarrow15x-30x=-24+20\)
\(\Leftrightarrow-15x=-4\)
\(\Rightarrow x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{4}{15}\)
b) \(\dfrac{2}{3}\cdot\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\dfrac{1}{2}=\dfrac{2}{3}-x\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{2}{9}+\dfrac{1}{2}=\dfrac{2}{3}-x\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{5}{18}=\dfrac{2}{3}-x\)
\(\Leftrightarrow6x+5=12-18x\)
\(\Leftrightarrow6x+18x=12-5\)
\(\Leftrightarrow24x=7\)
\(\Rightarrow x=\dfrac{7}{24}\)
Vậy \(x=\dfrac{7}{24}\)
\(a,0,5\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}x-x=\dfrac{-4}{5}+\dfrac{1}{6}+\dfrac{1}{2}\\ x\left(\dfrac{1}{2}-1\right)=\dfrac{-24}{30}+\dfrac{5}{30}+\dfrac{15}{30}\\ \dfrac{-1}{2}x=\dfrac{-2}{15}\\ x=\dfrac{-2}{15}:\dfrac{-1}{2}\\ x=\dfrac{-2}{15}\cdot\left(-2\right)\\ x=\dfrac{4}{15}\)
\(b,\dfrac{2}{3}\cdot\left(\dfrac{1}{2}\cdot x-\dfrac{1}{3}\right)+\dfrac{1}{2}=\dfrac{2}{3}-x\\ \dfrac{1}{3}x-\dfrac{2}{9}+\dfrac{1}{2}=\dfrac{2}{3}-x\\ \dfrac{1}{3}x+x=\dfrac{2}{3}+\dfrac{2}{9}-\dfrac{1}{2}\\ x\left(\dfrac{1}{3}+1\right)=\dfrac{12}{18}+\dfrac{4}{18}-\dfrac{9}{18}\\ \dfrac{4}{3}x=\dfrac{7}{18}\\ x=\dfrac{7}{18}:\dfrac{4}{3}\\ x=\dfrac{7}{18}\cdot\dfrac{3}{4}\\ x=\dfrac{7}{24}\)
MÌNH GHI KẾT QUẢ THÔI NHÉ
A,2
b,\(\dfrac{-2}{3}\)hoặc2
c,\(\dfrac{13}{15}\)hoặc\(\dfrac{17}{-15}\)
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}.\left(-14\right)=-6\)
c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)
\(=-10:\left(-\dfrac{5}{7}\right)\)
\(=14\)
d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)
\(=0:\dfrac{4}{5}\)
\(=0\)
a,
\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1-0,5=1,5\)
b,
\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)
c,
\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)
d,
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)
\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.