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a)\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(=\left(\frac{-1}{2}-\frac{3}{5}\right):\left(-3\right)+\frac{1}{6}:\left(-2\right)\)
\(=\frac{-11}{30}:\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(=\frac{11}{30}+\frac{1}{3}+\frac{-1}{12}\)
\(=\frac{37}{60}\)
b)\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(=\left(\frac{2}{25}-\frac{126}{125}\right):\frac{4}{7}:\left[\left(\frac{13}{4}-\frac{59}{9}\right).\frac{36}{17}\right]\)
\(=\frac{-116}{125}:\frac{4}{7}:\left[\frac{-119}{36}.\frac{36}{17}\right]\)
\(=\frac{-116}{125}:\frac{4}{7}:-7\)
\(=\frac{29}{125}\)
a)=\(\frac{1}{9}.\frac{2}{145}-\frac{13}{3}.\frac{2}{145}+\frac{2}{145}\)
=\(\frac{2}{145}\left(\frac{1}{9}-\frac{13}{3}+1\right)\)
=\(\frac{2}{145}.\frac{-29}{9}\)
=\(\frac{-2}{45}\)
Học tốt nha!!!^^
mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)
- \(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(=\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)+\frac{11}{125}\)
\(=-1+\frac{1}{2}+\frac{11}{125}\)
\(=-1+\frac{147}{125}\)
\(=\frac{22}{125}\)
2. \(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
\(=\left(1+2+3+4-3-2-1\right)\)\(+\left(-\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(-\frac{3}{4}-\frac{1}{4}\right)\)
\(=4-1-1-1\)
\(=1\)
a: \(=\dfrac{11}{125}-\dfrac{17}{18}+\dfrac{8}{18}-\dfrac{10}{14}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)
b: \(=\left(1+2+3+4-1\right)+\left(-\dfrac{1}{2}-\dfrac{2}{3}\right)=9-\dfrac{5}{6}=\dfrac{49}{6}\)
c: \(A=26:\left(6+\dfrac{1}{2}\right)=26:\dfrac{13}{2}=4\)