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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!!!^^
Đây là tính hợp lí ... mà câu a là 27,5 chứ không phải 2,75...
\(A=\dfrac{7,5-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{27,5-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{15}{2}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{55}{2}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{3}}\\ =\dfrac{3\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
b: \(=26:\left[\dfrac{3:0.1}{2.5\cdot2}+\dfrac{0.25\cdot4}{2}\right]+\dfrac{2}{3}\cdot\dfrac{21}{4}\)
\(=26:\left[\dfrac{30}{5}+1\right]+\dfrac{42}{12}\)
\(=\dfrac{26}{7}+\dfrac{42}{12}=\dfrac{101}{14}\)
c: \(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{\left(\dfrac{1}{34}+\dfrac{33}{34}\right)}{\dfrac{1}{2}+\dfrac{9}{2}}=1:5=\dfrac{1}{5}\)
Bài 1: Tìm x
a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)
\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)
Vậy: x=4
b) Ta có: \(\left|x\right|=-1.2\)
mà \(\left|x\right|\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
Bài 2: Tính
a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)
\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)
\(=-7.9\cdot10=-79\)
b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)
\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)
\(=3\cdot\dfrac{13}{3}=13\)
Bài 1:
a) 0,24 = 6/25
b) 0,245 = 49/200
c) 2,5324 = 5/2
d) 0,5 = 1/2
a) \(0,\left(24\right)=\frac{24}{99}=\frac{8}{33}\)
b)\(0,2\left(45\right)=\frac{245-2}{990}=\frac{243}{990}=\frac{27}{110}\)
c)\(2,5\left(324\right)=2+0,5\left(324\right)=2+\frac{5324-5}{9990}=2+\frac{197}{370}=\frac{937}{370}\)
d) \(0,5\left(3\right)=\frac{53-5}{90}=\frac{48}{90}=\frac{8}{15}\)
Bài 2 : \(M=\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(M=\frac{\frac{1}{2}+\frac{1}{3}-\frac{16-1}{90}}{\frac{5}{2}+\frac{5}{3}-\frac{83-8}{90}}\)
\(M=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{6}}=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)}=\frac{1}{5}\)
a)\(\frac{7}{3}.\left( { - 2,5} \right).\frac{6}{7} = \frac{7}{3}.\frac{6}{7}.\left( { - 2,5} \right) = 2.\left( { - 2,5} \right) = - 5\)
b)
\(\begin{array}{l}0,8.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9} - 0,2\\ = \frac{4}{5}.\frac{{ - 2}}{9} - \frac{4}{5}.\frac{7}{9}-\frac{2}{10}\\ = \frac{4}{5}.\left( {\frac{{ - 2}}{9} - \frac{7}{9}} \right) -\frac{1}{5}\\ = \frac{4}{5}.\left( { - 1} \right)-\frac{1}{5} \\= \frac{{ - 4}}{5}-\frac{1}{5}\\=\frac{-5}{5}\\=-1.\end{array}\)
1000+26