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a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
a) \(\left(\frac{-1}{6}+\frac{5}{-12}\right)+\frac{7}{12}=\left(\frac{-2}{12}+\frac{-5}{12}\right)+\frac{7}{12}=\left(\frac{-7}{12}\right)+\frac{7}{12}=0\)
b)\(\frac{7}{36}-\frac{8}{-9}+\frac{-2}{3}=\frac{7}{36}+\frac{32}{36}-\frac{24}{36}=\frac{15}{36}=\frac{5}{12}\)
c) \(\frac{3}{5}-\frac{2}{5}.\frac{10}{12}=\frac{3}{5}-\frac{2}{5}.\frac{5}{6}=\frac{3}{5}-\frac{1}{3}=\frac{9}{15}-\frac{5}{15}=\frac{4}{15}\)
d) \(\frac{2}{\left(-3\right)^2}+\frac{5}{-13}-\frac{-3}{4}=\frac{2}{9}-\frac{5}{13}+\frac{3}{4}=\frac{8}{36}-\frac{15}{36}+\frac{27}{36}=\frac{5}{9}\)
a) \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
= \(\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
= \(0:\frac{3}{7}\)
= \(0\)
b) \(\frac{2}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)
= \(\frac{1}{4}:\frac{1}{6}+\frac{7}{8}:\frac{-7}{18}\)
=\(\frac{1}{4}.6+\frac{7}{8}.\frac{-18}{7}\)
= \(\frac{3}{2}-\frac{3}{4}\)
= \(\frac{3}{4}\)
a)\(=\frac{13}{4}-\frac{1}{3}\)
\(=\frac{35}{12}\)
b)\(=\left(\frac{7}{6}-\frac{1}{2}\right).\frac{2}{3}-\frac{1}{4}\)
\(=\frac{2}{3}.\frac{2}{3}-\frac{1}{4}\left(=1-\frac{1}{4}\right)\)
\(=\frac{4}{9}-\frac{1}{4}\left(=\frac{3}{4}\right)\)
\(=\frac{7}{36}\)
a, \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)
\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)
\(A=\frac{-7}{20}+\frac{1}{2}\)
\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)
b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)
\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)
\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)
\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)
\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)
c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)
\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)
\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)
\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)
\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)
d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)
\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a. \(\frac{2}{5}+\frac{3}{5}:\left(\frac{3}{5}+\frac{-2}{3}\right)-3\frac{1}{2}\)
\(=\frac{2}{5}+\frac{3}{5}:\frac{-1}{15}-\frac{7}{2}\)
\(=\frac{2}{5}+\left(-9\right)-\frac{7}{2}\)
\(=-\frac{43}{5}-\frac{7}{2}\)
\(=-\frac{121}{10}\)
\(a)\frac{2}{5}+\frac{3}{5}:\left(\frac{3}{5}+\frac{-2}{3}\right)-3\frac{1}{2}\)
\(=\frac{2}{5}+\frac{3}{5}:\frac{-1}{15}-\frac{7}{2}\)
\(=\frac{2}{5}+(-9)-\frac{7}{2}\)
\(=-\frac{43}{5}-\frac{7}{2}=-\frac{121}{10}\)