Tìm các số tự nhiên a,b thỏa mãn điều kiện: \(\dfrac{14}{17}< \dfrac{a}{b}< \dfrac{23}{29}\) và 8b-9a=31
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Lời giải:
Ta có:
\(\frac{x-2}{5}=\frac{1-x}{6}\)
$\Rightarrow 6(x-2)=5(1-x)$
$\Rightarrow 6x-12=5-5x$
$\Rightarrow 11x=17$
$\Rightarrow x=\frac{17}{11}$
6.1
\(\frac{3}{8}-\frac{4}{5}-\frac{-17}{40}=\frac{15}{40}-\frac{32}{40}+\frac{17}{40}=\frac{15+17-32}{40}=0\)
6.2
\(\frac{3}{4}-\frac{16}{32}+\frac{4}{-3}=\frac{9}{12}-\frac{6}{12}-\frac{16}{12}=\frac{9-6-16}{12}=\frac{-13}{12}\)
6.3
\(\frac{-4}{7}+\frac{2}{3}.\frac{-9}{14}=\frac{-4}{7}-\frac{3}{7}=-\frac{7}{7}=-1\)
6.4
\(8\frac{2}{7}-(3\frac{4}{9}+4\frac{2}{7})=8+\frac{2}{7}-(7+\frac{4}{9}+\frac{2}{7})=1-\frac{4}{9}=\frac{5}{9}\)
6.5
\((\frac{2}{3}-1\frac{1}{2}):\frac{4}{3}+\frac{1}{2}=\frac{-5}{6}.\frac{3}{4}+\frac{1}{2}=\frac{-5}{8}+\frac{1}{2}=\frac{-5}{8}+\frac{4}{8}=\frac{-1}{8}\)
6.6
\(\frac{-5}{13}+\frac{2}{5}+\frac{-8}{13}+\frac{3}{5}-\frac{3}{7}\\ =(\frac{-5}{13}+\frac{-8}{13})+(\frac{2}{5}+\frac{3}{5})-\frac{3}{7}\\ =-1+1-\frac{3}{7}=\frac{-3}{7}\)
Bài 8:
a/ $=(\frac{2}{3}-\frac{2}{3})+\frac{5}{7}=0+\frac{5}{7}=\frac{5}{7}$
b/ $=(\frac{5}{13}+\frac{8}{13})+\frac{-5}{7}+(\frac{-20}{41}+\frac{-21}{41})$
$=1-\frac{5}{7}-1=-\frac{5}{7}$
c/ $=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}$
$=\frac{1}{2}(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+\frac{13-11}{11.13})$
$=\frac{1}{2}(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13})$
$=\frac{1}{2}(\frac{1}{3}-\frac{1}{13})$
$=\frac{5}{39}$
b/
$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x(x+1)}=\frac{2021}{2022}$
$\Rightarrow \frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x(x+1)}=\frac{2021}{2022}$
$\Rightarrow 2(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x(x+1)})=\frac{2021}{2022}$
$\Rightarrow 2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1})=\frac{2021}{2022}$
$\Rightarrow 2(\frac{1}{2}-\frac{1}{x+1})=\frac{2021}{2022}$
$\Rightarrow 1-\frac{2}{x+1}=\frac{2021}{2022}$
$\Rightarrow \frac{2}{x+1}=\frac{1}{2022}$
$\Rightarrow x+1=4044$
$\Rightarrow x=4043$
Lời giải:
a/
$(x+\frac{4}{1}-\frac{1}{3}):(2+\frac{1}{6}-\frac{1}{4})=\frac{7}{46}$
$(x+\frac{11}{3}):\frac{23}{12}=\frac{7}{46}$
$x+\frac{11}{3}=\frac{7}{46}.\frac{23}{12}=\frac{7}{24}$
$x=\frac{7}{24}-\frac{11}{3}=\frac{-27}{8}$
Lời giải:
$8b-9a=31\Rightarrow a=\frac{8b-31}{9}$
$\frac{a}{b}> \frac{14}{17}$
$\Rightarrow 17a-14b>0$
$\Rightarrow 17.\frac{8b-31}{9}-14b>0$
$\Rightarrow 17(8b-31)-126b>0$
$\Rightarrow 10b> 527\Rightarrow b> 52,7(1)$
Lại có:
$\frac{a}{b}< \frac{23}{29}$
$\Rightarrow 29a-23b<0$
$\Rightarrow 29.\frac{8b-31}{9}-23b<0$
$\Rightarrow 29(8b-31)-207b<0$
$\Rightarrow 25b< 899$
$\Rightarrow b< 35,96(2)$
Từ $(1); (2)\Rightarrow 52,7< 35,96$ (vô lý)
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