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a)\(\dfrac{-3}{29}+\dfrac{16}{58}\)\(=\dfrac{-3}{29}+\dfrac{8}{29}=\dfrac{5}{29}\)
b) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{-3}{5}\)
c) \(\dfrac{-8}{18}+\dfrac{-15}{27}=\dfrac{-4}{9}+\dfrac{-5}{9}=\dfrac{-9}{9}=-1\)
a) \(\dfrac{-3}{29}+\dfrac{16}{58}=\dfrac{-3}{29}+\dfrac{8}{29}=\dfrac{-3+8}{29}=\dfrac{5}{29}\)
b) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{1+\left(-4\right)}{5}=\dfrac{-3}{5}\)
c) \(\dfrac{-8}{18}+\dfrac{-15}{27}=\dfrac{-4}{9}+\dfrac{-5}{9}=\dfrac{-4+\left(-5\right)}{9}=\dfrac{-9}{9}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính các tổng dưới đây sau khi đã rút gọn phân số :
a)\(\dfrac{7}{21}\) + \(\dfrac{9}{-36}\) = \(\dfrac{7}{21}\)+\(\dfrac{-9}{36}\)=\(\dfrac{1}{3}\)+\(\dfrac{-1}{4}\)=\(\dfrac{4}{12}\)+\(\dfrac{-3}{12}\)=\(\dfrac{1}{12}\)
b) \(\dfrac{-12}{18}\)+\(\dfrac{-21}{35}\)=\(\dfrac{-2}{3}\)+\(\dfrac{-3}{5}\)=\(\dfrac{-10}{15}\)+\(\dfrac{-9}{15}\)=\(\dfrac{-19}{15}\)
c) \(\dfrac{-3}{21}\)+\(\dfrac{6}{42}\)=\(\dfrac{-1}{7}\)+\(\dfrac{1}{7}\)=0
d) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
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\(a,\dfrac{7}{35},\dfrac{18}{54},\dfrac{-15}{125},\dfrac{-4}{25}\)
Các thừa số đã tối giản : \(\dfrac{-4}{25}\)
\(\dfrac{7}{35}=\dfrac{7:7}{35:7}=\dfrac{1}{5}\) , \(\dfrac{18}{54}=\dfrac{18:18}{54:18}=\dfrac{1}{3}\)
\(\dfrac{-15}{125}=\dfrac{-15:5}{125:5}=\dfrac{-3}{25}\)
\(b,\dfrac{27}{45},\dfrac{21}{28},\dfrac{8}{14},\dfrac{18}{-60},\dfrac{-270}{360}\)
Các thừa số đã tối giản là : ko có
\(\dfrac{27}{45}=\dfrac{27:9}{45:9}=\dfrac{3}{5}\) , \(\dfrac{21}{28}=\dfrac{21:7}{28:7}=\dfrac{3}{4}\)
\(\dfrac{8}{14}\)\(=\dfrac{8:2}{14:2}=\dfrac{4}{7}\) , \(\dfrac{18}{-60}=\dfrac{18:6}{-60:6}=\dfrac{3}{-10}=\dfrac{-3}{10}\)
\(\dfrac{-270}{360}=\dfrac{-270:90}{360:90}=\dfrac{-3}{4}\)
\(c,\dfrac{3.4+3.7}{6.5+9}\) = \(\dfrac{3.\left(4+7\right)}{30+9}\) = \(\dfrac{3.11}{39}\) = \(\dfrac{3.11}{3.13}=\dfrac{11}{13}\)
\(\dfrac{-63}{81},\dfrac{9.6}{9.35},\dfrac{7.2+8}{2.14.5}\)
Các p/s đã tối giản : ko có
\(\dfrac{-63}{81}=\dfrac{-63:9}{81:9}=\dfrac{-7}{9}\) , \(\dfrac{9.6}{9.35}=\dfrac{6}{35}\)
\(\dfrac{7.2+8}{2.14.5}=\dfrac{14+8}{28.5}=\dfrac{22}{140}=\dfrac{11}{70}\)
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a. \(\dfrac{-3}{5}\)
b. \(\dfrac{-2}{3}\) c. \(\dfrac{4}{39}\) d. \(\dfrac{26}{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{-5}{21}+\dfrac{-6}{42}=\dfrac{-5}{21}+\dfrac{-3}{21}=\dfrac{\left(-5\right)+\left(-3\right)}{21}=\dfrac{-8}{21}\)
\(b,\dfrac{16}{40}+\dfrac{-27}{45}=\dfrac{2}{5}+\dfrac{-3}{5}=\dfrac{2+\left(-3\right)}{5}=\dfrac{-1}{5}\)
\(c,\dfrac{-4}{18}+\dfrac{-12}{30}=\dfrac{-2}{9}+\dfrac{-2}{5}=\dfrac{-10}{45}+\dfrac{-18}{45}=\dfrac{\left(-10\right)+\left(-18\right)}{45}=\dfrac{-28}{45}\)
a)\(\dfrac{-5}{21}\)+\(\dfrac{-6}{42}\)= \(\dfrac{-10}{42}\)+\(\dfrac{-6}{42}\)= \(\dfrac{-16}{42}\)= \(\dfrac{-8}{21}\)
b)\(\dfrac{16}{40}\)+ \(\dfrac{-27}{45}\)= \(\dfrac{2}{5}\)+\(\dfrac{-3}{5}\)= \(\dfrac{-5}{5}\)= (-1)
c) \(\dfrac{-4}{18}\)+\(\dfrac{-12}{30}\)= \(\dfrac{-2}{9}\)+\(\dfrac{-2}{5}\)= \(\dfrac{-10}{45}\)+\(\dfrac{-18}{45}\)= \(\dfrac{-28}{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
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Giải:
Ta có:
\(\dfrac{a}{b}=\dfrac{-8}{11}\left(1\right)\Leftrightarrow1-\dfrac{a}{b}=1-\dfrac{-8}{11}\)
Hay \(\dfrac{b-a}{b}=\dfrac{11+8}{11}=\dfrac{19}{11}\left(2\right)\)
Thay \(b-a=190\) vào \(\left(2\right)\) ta được:
\(\dfrac{190}{b}=\dfrac{19}{11}\Leftrightarrow190.11=19b\Leftrightarrow b=110\)
Thay \(b=110\) vào \(\left(1\right)\) ta được:
\(\dfrac{a}{110}=\dfrac{-8}{11}\Leftrightarrow11a=-8.110\Leftrightarrow a=-80\)
Vậy phân số \(\dfrac{a}{b}\) cần tìm là \(\dfrac{-80}{110}\)
Thay b - a = 190 vào (1) ta được:
Phân số a/b phải tìm là -80/110
\(-\dfrac{2}{3}\)
-2/3