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\(\begin{array}{l}a)\frac{{17}}{{11}} - \left( {\frac{6}{5} - \frac{{16}}{{11}}} \right) + \frac{{26}}{5}\\ = \frac{{17}}{{11}} - \frac{6}{5} + \frac{{16}}{{11}} + \frac{{26}}{5}\\ = (\frac{{17}}{{11}} + \frac{{16}}{{11}}) + (\frac{{26}}{5} - \frac{6}{5})\\ = \frac{{33}}{{11}} + \frac{{20}}{5}\\ = 3 + 4\\ = 7\\b)\frac{{39}}{5} + \left( {\frac{9}{4} - \frac{9}{5}} \right) - \left( {\frac{5}{4} + \frac{6}{7}} \right)\\ = \frac{{39}}{5} + \frac{9}{4} - \frac{9}{5} - \frac{5}{4} - \frac{6}{7}\\ = (\frac{{39}}{5} - \frac{9}{5}) + (\frac{9}{4} - \frac{5}{4}) - \frac{6}{7}\\ = \frac{{30}}{5} + \frac{4}{4} - \frac{6}{7}\\ = 6 + 1 - \frac{6}{7}\\ = 7 - \frac{6}{7}\\ = \frac{{49}}{7} - \frac{6}{7}\\ = \frac{{43}}{7}\end{array}\)
mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?
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)\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(=\left(\frac{2}{25}-\frac{126}{125}\right):\frac{4}{7}:\left[\left(\frac{13}{4}-\frac{59}{9}\right).\frac{36}{17}\right]\)
\(=\frac{-116}{125}:\frac{4}{7}:\left[\frac{-119}{36}.\frac{36}{17}\right]\)
\(=\frac{-116}{125}:\frac{4}{7}:-7\)
\(=\frac{29}{125}\)
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)
a)\(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{7}{14}\)
=\(\frac{11}{125}-\left(\frac{17}{18}-\frac{4}{9}\right)+\left(\frac{17}{14}-\frac{5}{7}\right)\)
=\(\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
=\(\frac{11}{125}+\left(\frac{1}{2}-\frac{1}{2}\right)\)
=\(\frac{11}{125}\)
b)................................đề bài......................................................
= ( 1 - 1 ) + ( 2 - 2 ) + ( 3 - 3 ) + ( -1/2 + -1/2 ) + ( -2/3 + -1/3) + ( -3/4 + -1/4) + 4
= 0 + 0 + 0 + (-1) + (-1) + (-1) + 4
= 1
a)|-10|:(-2):(-5)+(-3)2
=1+9
=10
b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)
=[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]
=(-1)+(-1)+(-1)+...+(-1)
Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)
Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)
c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)
\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)
\(=\frac{3}{4}\)
d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)
\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)
\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)
\(=-\frac{1099}{1955}\)
e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)
\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)
\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)
\(=-\frac{11}{4}\)
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: \(A=26:\left(6+\dfrac{1}{2}\right)=26:\dfrac{13}{2}=4\)