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a ) \(\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
= \(\frac{2}{5}+\frac{-1}{6}+\frac{-3}{4}+\frac{2}{3}\)
= \(\left(\frac{2}{5}+\frac{-3}{4}\right)+\left(\frac{-1}{6}+\frac{2}{3}\right)\)
= \(\left(\frac{8}{20}+\frac{-15}{20}\right)+\left(\frac{-1}{6}+\frac{4}{6}\right)\)
= \(\left(\frac{8+\left(-15\right)}{20}\right)+\left(\frac{\left(-1\right)+4}{6}\right)\)
= \(\frac{-7}{20}+\frac{1}{2}\)
= \(\frac{-7}{20}+\frac{10}{20}=\frac{\left(7\right)+10}{20}=\frac{3}{20}\)
tk mk nha
đang âm rất nhiều rồi , giúp nha !!!!!
\(\left(2+\frac{5}{6}\right)\div1\frac{1}{5}+\frac{-7}{12}\)
\(=\left(\frac{12}{6}+\frac{5}{6}\right)\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\times\frac{5}{6}-\frac{7}{12}\)
\(=\frac{85}{12}-\frac{7}{12}\)
\(=\frac{78}{12}=\frac{13}{2}\)
\(\left(15-6\frac{13}{18}\right)\div11\frac{1}{7}-2\frac{1}{8}\div1\frac{11}{40}\)
\(=9\frac{13}{18}\div\frac{78}{7}-\frac{17}{8}\div\frac{51}{40}\)
\(=\frac{175}{18}\div\frac{78}{7}-\frac{17}{8}\times\frac{40}{51}\)
\(=\frac{175}{18}\times\frac{7}{78}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{2340}{1404}\)
\(=\frac{-1115}{1404}\)
Câu a:
{3849 - 2.[146 - (2.14 + 6: 2)]} : 32
= {3849 - 2.[146 - (28 + 3)]} : 32
= {3489 - 2.[146 - 31]} : 32
= {3489 - 2.115} : 32
= {3489 - 230} : 32
= 3259 : 32
= 3259/32
Câu b:
134 - 2.{156 - 6[54 - 2.(9 + 6)]}
= 134 - 2.{156 - 6.[54- 2. 15]}
= 134 - 2.{156 - 6.[54 - 30]}
= 134 - 2.{156 - 6.24]
= 134 - 2.12
= 134 - 24
= 110
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1
(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)
(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)
\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)
(44\(x\) + 3) : \(\frac{11}{5}\) = - 4
44\(x\) + 3 = - 4 x \(\frac{11}{5}\)
44\(x\) + 3 = - \(\frac{44}{5}\)
44\(x\) = - \(\frac{44}{5}\) - 3
44\(x\) = - \(\frac{59}{5}\)
\(x=-\frac{59}{5}:44\)
\(x\) = - \(\frac{59}{250}\)
Vậy \(x\) = - \(\frac{59}{250}\)
a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)