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\(\dfrac{-5}{9}\)=\(\dfrac{-45}{b}\)
⇒ b= [9. (-45)] : -5
⇒ b= -405 : -5
⇒ b= 81
⇒ \(\dfrac{-45}{81}\)
\(\dfrac{a}{27}\)= \(\dfrac{-5}{9}\)
⇒ a= [ 27 .(-5) ] : 9
⇒ a= -135 : 9
⇒ a= -15
⇒ \(\dfrac{-15}{27}\)
⇒ \(\dfrac{-15}{27}\)=\(\dfrac{-5}{9}\)=\(\dfrac{-45}{81}\)
\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
⇒\(a=\dfrac{-5.27}{9}=-15\)
⇒\(b=\dfrac{-45.9}{5}=-81\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
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\(-\dfrac{x}{27}-1=\dfrac{2}{3}\)
\(\Rightarrow-\dfrac{x}{27}=\dfrac{2}{3}+1=\dfrac{5}{3}\)
\(\Rightarrow-3x=27.5\)
\(\Rightarrow x=-135:\left(-3\right)\)
\(\Rightarrow x=-45\)
`->B`
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\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+\dfrac{14}{15}+\dfrac{20}{21}+\dfrac{27}{28}+\dfrac{35}{36}+\dfrac{44}{45}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{21}\right)+\left(1-\dfrac{1}{28}\right)+\left(1-\dfrac{1}{36}\right)+\left(1-\dfrac{1}{45}\right)\\ =8-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\\ =8-\left(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ =8-\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\right)\\ =8-2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=8-2.\dfrac{2}{5}=8-\dfrac{4}{5}=\dfrac{36}{5}\)
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Lời giải:
a, \(\dfrac{12}{19}.\dfrac{7}{15}.\dfrac{-13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\dfrac{12}{19}.\dfrac{-7}{15}.\dfrac{13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\left(\dfrac{12}{19}.\dfrac{19}{12}\right).\left(\dfrac{13}{17}.\dfrac{17}{13}\right).\dfrac{-7}{15}\)
\(=1.1.\dfrac{-7}{15}\)
\(=\dfrac{-7}{15}\)
b, \(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{27}=\dfrac{-5}{9}\\\dfrac{-45}{b}=\dfrac{-5}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-5.27\\-5b=-45.9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-135\\-5b=-405\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-15\\b=-81\end{matrix}\right.\)
Vậy, \(a=-15;b=-81\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
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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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bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{-5}{9}\cdot27=-15\\b=\dfrac{-45\cdot9}{-5}=9\cdot9=81\end{matrix}\right.\)