`#3107.101107`
`1.`
`a)`
\(\dfrac{7}{23}-\dfrac{5}{14}+\dfrac{1}{2}-\dfrac{9}{14}+\dfrac{16}{23}\\ =\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\left(\dfrac{5}{14}+\dfrac{9}{14}\right)+\dfrac{1}{2}\\ =\dfrac{23}{23}-\dfrac{14}{14}+\dfrac{1}{2}\\ =1-1+\dfrac{1}{2}\\ =\dfrac{1}{2}\)
`b)`
\(\left(-\dfrac{2}{3}\right)\cdot\dfrac{3}{11}+\left(-\dfrac{16}{9}\right)\cdot\dfrac{3}{11}\\ =\dfrac{3}{11}\cdot\left(-\dfrac{2}{3}-\dfrac{16}{9}\right)\\ =\dfrac{3}{11}\cdot\left(-\dfrac{22}{9}\right)\\ =-\dfrac{2}{3}\)
`2.`
`a)`
\(-\dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{1}{3}\\ \Rightarrow-\dfrac{3}{7}x=\dfrac{1}{6}\\ \Rightarrow x=\dfrac{1}{6}\div\left(-\dfrac{3}{7}\right)\\ \Rightarrow x=-\dfrac{7}{18}\)
`b)`
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
_____
`1.`
`a)`
\(-\dfrac{9}{17}+6,72+\dfrac{-8}{17}+\left(-4,72\right)\\ =\left(-\dfrac{9}{17}-\dfrac{8}{17}\right)+\left(6,72-4,72\right)\\ =-\dfrac{17}{17}+2\\ =-1+2=1\)
`b)`
\(2\dfrac{1}{5}-\left(-\dfrac{3}{4}+\dfrac{1}{5}\right)\\ =\dfrac{11}{5}+\dfrac{3}{4}-\dfrac{1}{5}\\ =\left(\dfrac{11}{5}-\dfrac{1}{5}\right)+\dfrac{3}{4}\\ =\dfrac{10}{5}-\dfrac{3}{4}\\ =2-\dfrac{3}{4}\\ =\dfrac{5}{4}\)
`c)`
\(\left(-\dfrac{1}{3}\right)^3-\dfrac{3}{8}\div \left(\dfrac{1}{2}\right)^3-\dfrac{5}{2}\cdot\left(-2\right)\\ =\left(-\dfrac{1}{27}\right)-\dfrac{3}{8}\div\dfrac{1}{8}-\left(-5\right)\\ =-\dfrac{1}{27}-3+5\\ =-\dfrac{1}{27}+2\\ =\dfrac{53}{27}\)
`d)`
\(4,1\cdot\dfrac{-5}{12}-6,2+4,1\cdot\dfrac{-7}{12}\\ =4,1\cdot\left(-\dfrac{5}{12}-\dfrac{7}{12}\right)-6,2\\ =4,1\cdot\left(-\dfrac{12}{12}\right)-6,2\\ 4,1\cdot\left(-1\right)-6,2\\ =-4,1-6,2\\ =-10,3\)
`2.`
`a)`
\(x+\left(-\dfrac{1}{9}\right)=-\dfrac{7}{6}\\ \Rightarrow x=-\dfrac{7}{6}-\left(-\dfrac{1}{9}\right)\\ \Rightarrow x=-\dfrac{7}{6}+\dfrac{1}{9}\\ \Rightarrow x=-\dfrac{19}{18}\)
`b)`
\(\left(x-\dfrac{4}{7}\right)\div\dfrac{-5}{3}=0,2\\ \Rightarrow x-\dfrac{4}{7}=0,2\cdot\left(-\dfrac{5}{3}\right)\\ \Rightarrow x-\dfrac{4}{7}=-\dfrac{1}{3}\\ \Rightarrow x=-\dfrac{1}{3}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{5}{21}\)
`c)`
\(\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{8}\\ \Rightarrow\left(2x-\dfrac{1}{3}\right)^3=\left(\dfrac{1}{2}\right)^3\\ \Rightarrow2x-\dfrac{1}{3}=\dfrac{1}{2}\\ \Rightarrow2x=\dfrac{1}{2}+\dfrac{1}{3}\\ \Rightarrow2x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}\div2\\ \Rightarrow x=\dfrac{5}{12}\)
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`13.`
`a)`
Ta có: \(\dfrac{139}{303}>\dfrac{138}{303}=\dfrac{46}{101}\)
Mà \(\dfrac{35}{101}< \dfrac{46}{101}\)
\(\Rightarrow\dfrac{139}{303}>\dfrac{35}{101}\)
`b)`
Ta có:
\(\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^{8-2}=\left(\dfrac{1}{2}\right)^6\)
\(\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3=\left(\dfrac{1}{2}\right)^{3+3}=\left(\dfrac{1}{2}\right)^6\)
Vì \(\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^6\\ \Rightarrow\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3\)
`14.`
`a)`
\(-\dfrac{11}{24}+\dfrac{3}{4}-\dfrac{13}{24}\\ =\left(-\dfrac{11}{24}-\dfrac{13}{24}\right)+\dfrac{3}{4}\\ =-\dfrac{24}{24}+\dfrac{3}{4}\\ =-1+\dfrac{3}{4}\\ =-\dfrac{1}{4}\)
`b)`
\(-\dfrac{5}{9}-\left(\dfrac{8}{15}+\dfrac{4}{9}\right)+\dfrac{7}{15}\\ =-\dfrac{5}{9}-\dfrac{8}{15}-\dfrac{4}{9}+\dfrac{7}{15}\\ =\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)-\left(\dfrac{8}{15}+\dfrac{7}{15}\right)\\ =-\dfrac{9}{9}-\dfrac{15}{15}\\ =-1-1=-2\)
`c)`
\(\dfrac{5}{9}\div2,4-\dfrac{41}{9}\div2,4\\ =\dfrac{5}{9}\div\dfrac{12}{5}-\dfrac{41}{9}\div\dfrac{12}{5}\\ =\dfrac{5}{9}\cdot\dfrac{5}{12}-\dfrac{41}{9}\cdot\dfrac{5}{12}\\ =\dfrac{5}{12}\cdot\left(\dfrac{5}{9}-\dfrac{41}{9}\right)\\ =\dfrac{5}{12}\cdot\left(-\dfrac{36}{9}\right)\\ =\dfrac{5}{12}\cdot\left(-4\right)\\ =-\dfrac{5}{3}\)
`a)`
`b)`
`c)`
Câu 15:
a) \(x\cdot\left(-\dfrac{3}{7}\right)=\dfrac{5}{21}\)
\(x=\dfrac{5}{21}:\left(-\dfrac{3}{7}\right)\)
\(x=\dfrac{5}{21}\cdot\left(\dfrac{-7}{3}\right)\)
\(x=\dfrac{-5}{9}\)
b) \(x-\dfrac{1}{2}=3\dfrac{1}{2}:\dfrac{2}{7}\)
\(x-\dfrac{1}{2}=\dfrac{7}{2}:\dfrac{2}{7}\)
\(x-\dfrac{1}{2}=\dfrac{49}{4}\)
\(x=\dfrac{49}{4}+\dfrac{1}{2}\)
\(x=\dfrac{51}{4}\)
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