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\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
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 \(Q=\dfrac{29}{125}\)
\(a,A=\dfrac{7}{35}+\left(-1\dfrac{3}{4}+\dfrac{12}{7}\right)-\left(\dfrac{1}{4}-\dfrac{2}{7}-\dfrac{12}{35}\right)-\dfrac{3}{7}\)\(A=\dfrac{7}{35}-\dfrac{7}{4}+\dfrac{12}{7}-\dfrac{1}{4}+\dfrac{2}{7}+\dfrac{13}{35}-\dfrac{3}{7}\\ A=\left(\dfrac{7}{35}+\dfrac{13}{35}\right)-\left(\dfrac{7}{4}-\dfrac{1}{4}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}-\dfrac{3}{7}\right)\)
\(A=\dfrac{4}{7}-\dfrac{3}{2}+\dfrac{11}{7}\\ A=\left(\dfrac{4}{7}+\dfrac{11}{7}\right)-\dfrac{3}{2}\\ A=\dfrac{15}{7}-\dfrac{3}{2}=\dfrac{9}{14}\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)
= \(-\dfrac{73}{10}\) : - 5
= \(\dfrac{73}{50}\)
b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)
= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)
= \(\dfrac{2}{41}\)