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Bài 1 :
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{17}+\dfrac{3}{37}}{\dfrac{5}{7}-\dfrac{5}{17}+\dfrac{5}{37}}+\dfrac{2}{5}=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{17}+\dfrac{1}{37}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{17}+\dfrac{1}{37}\right)}+\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{5}{5}=1\)
\(\text{Câu 1 :}\)
\(A=\dfrac{5}{17}+\dfrac{-4}{9}-\dfrac{20}{31}+\dfrac{12}{17}-\dfrac{11}{31}\\ A=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{20}{31}+\dfrac{11}{31}\right)+\dfrac{-4}{9}\\ A=1-1+-\dfrac{4}{9}\\ A=-\dfrac{4}{9}\)
\(B=\dfrac{-3}{7}+\dfrac{7}{15}+\dfrac{-4}{7}+\dfrac{8}{15}-\dfrac{-2}{3}\\ B=\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{7}{15}+\dfrac{8}{18}\right)-\dfrac{-2}{3}\\ B=\left(-1\right)+1+\dfrac{2}{3}\\ B=\dfrac{2}{3}\)
\(\text{Câu 2 : }\)
\(A< \dfrac{x}{9}\le B\\ \Rightarrow\dfrac{-4}{9}< \dfrac{x}{9}\le\dfrac{2}{3}\\ \Rightarrow\dfrac{-4}{9}< \dfrac{x}{9}\le\dfrac{6}{9}\\ \Rightarrow-4< x\le6\\ \Rightarrow x\in\left\{\pm4;\pm3;\pm2;\pm1;0;5;6\right\}\)
Mk nhầm chút nhé..
x không bằng -4 nhé. Nếu x bằng -4 thì bài sẽ như thế này:
\(-4\le x\le6\)
\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)
\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)
\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)
\(=-\dfrac{304}{189}\)
\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)
\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)
\(=\dfrac{-22489}{7680}\)
a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)
= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)
= 0 + \(\dfrac{11}{125}\)
= \(\dfrac{11}{125}\)
b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +
\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4
= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4
= -1
c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)
= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)
= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)
= \(\dfrac{-7}{75}\)
d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)
= 1 + (-1)
= 0
a, \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)=\dfrac{2}{3}-4.\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b, \(-\dfrac{2}{3}.\dfrac{3}{11}+\dfrac{-16}{9}.\dfrac{3}{11}=\dfrac{3}{11}.\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}.\dfrac{-22}{9}=-\dfrac{2}{3}\)
Câu c lấy máy tính tính nha!
c,
c) \(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{10}{14}+\dfrac{8}{18}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{17}{18}+\dfrac{8}{18}\right)+\left(-\dfrac{10}{14}+\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}+\left(-\dfrac{9}{18}\right)+\dfrac{7}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}+0=\dfrac{11}{125}\)
a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)
= \(\dfrac{3}{11}\)
b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
= \(\dfrac{3}{5}\)
c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)
= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)
= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)
= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)