\(\dfrac{xx}{9}-\dfrac{5}{yy}=\dfrac{25}{14}\)
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a) \(\dfrac{3}{5}+\dfrac{7}{25}=\dfrac{15}{25}+\dfrac{7}{25}=\dfrac{15+7}{25}=\dfrac{22}{25}\)
b) \(\dfrac{8}{11}-\dfrac{19}{33}=\dfrac{24}{33}-\dfrac{19}{33}=\dfrac{24-19}{33}=\dfrac{5}{33}\)
c) \(\dfrac{16}{21}\times\dfrac{3}{5}=\dfrac{16\times3}{21\times5}=\dfrac{48}{105}=\dfrac{16}{35}\)
d) \(\dfrac{14}{41}\div\dfrac{7}{9}=\dfrac{14}{41}\times\dfrac{9}{7}=\dfrac{14\times9}{41\times7}=\dfrac{126}{287}=\dfrac{18}{41}\)
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a)\(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}\\ =\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}\\ =\dfrac{14-15+1}{35}\\ =\dfrac{0}{35}=0\)
b)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\\ =\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}\\ =-1+1+\dfrac{2}{7}\\ =0+\dfrac{2}{7}=\dfrac{2}{7}\)
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a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A = - 2 + (-1)\\A = - 3\end{array}\)
b,
Cách 1:
\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)
Cách 2:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B = \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)
s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)
\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)
t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)
\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)
\(14:\dfrac{2}{11}=14\cdot\dfrac{11}{2}=77\)
\(b.\)
\(\dfrac{25}{14}:\dfrac{5}{7}=\dfrac{25}{14}\cdot\dfrac{7}{5}=\dfrac{5}{2}\)
\(c.\)
\(-\dfrac{10}{27}:\dfrac{5}{9}=\dfrac{-10}{27}\cdot\dfrac{9}{5}=-\dfrac{2}{3}\)
\(d.\)
\(90:30\%=\dfrac{90\cdot100}{30}=300\)
a, \(14:\dfrac{2}{11}=14.\dfrac{11}{2}=77\)
b, \(\dfrac{25}{14}:\dfrac{5}{7}=\dfrac{25}{14}.\dfrac{7}{5}=\dfrac{5}{2}\)
c, \(\dfrac{-10}{27}:\dfrac{5}{9}=\dfrac{-10}{27}.\dfrac{9}{5}=\dfrac{-2}{3}\)
d, \(90:30\%=90:\dfrac{30}{100}=90.\dfrac{100}{30}=300\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 9/18 - (-7/12) + 13/32
= 13/12 + 13/32
= 143/96
b) (5/-8) + 14/25 - 6/10
= (-13/200) - 6/10
= -133/200
Chúc bạn học tốt!! ^^
a: \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)
\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)
\(=\dfrac{48}{96}+\dfrac{56}{96}+\dfrac{39}{96}\)
\(=\dfrac{143}{96}\)
b: \(\dfrac{-5}{8}+\dfrac{14}{25}-\dfrac{6}{10}\)
\(=\dfrac{-125}{200}+\dfrac{112}{200}-\dfrac{120}{200}\)
\(=\dfrac{-133}{200}\)