Bài 1:
a: \(\dfrac{2}{9}-\dfrac{7}{8}x=\dfrac{1}{3}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}\)
hay \(x=\dfrac{-1}{9}:\dfrac{7}{8}=\dfrac{-1}{9}\cdot\dfrac{8}{7}=\dfrac{-8}{63}\)
b: \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{4}{3}:x=\dfrac{4}{7}\)
hay \(x=\dfrac{3}{7}\)
Bài 2:
a: Ta có: \(A=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{37\cdot39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{4}{13}\)
b: Ta có: \(B=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{32\cdot35}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\)
\(=\dfrac{1}{2}-\dfrac{1}{35}=\dfrac{33}{70}\)
Bài 1: Tìm x biết
a) \(\dfrac{2}{9}-\dfrac{7}{8}.x=\dfrac{1}{3}\) b) \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
\(\dfrac{7}{8}.x=\dfrac{2}{9}-\dfrac{1}{3}\) \(\dfrac{4}{3}:x=\dfrac{1}{7}-\dfrac{5}{7}\)
\(\dfrac{7}{8}.x=\dfrac{-1}{9}\) \(\dfrac{4}{3}:x=\dfrac{-4}{7}\)
\(x=\dfrac{-1}{9}:\dfrac{7}{8}\) \(x=\dfrac{4}{3}:\dfrac{-4}{7}\)
Vậy \(x=\dfrac{-8}{63}\) Vậy \(x=\dfrac{-7}{3}\)
c) \(\dfrac{5}{3}.x-\dfrac{1}{4}=\dfrac{2}{6}\) d) \(\dfrac{13}{4}-\dfrac{11}{5}:\dfrac{2}{10}+\dfrac{3}{4}\le x< \dfrac{5}{14}:\dfrac{15}{14}+1\)
\(\dfrac{5}{3}.x=\dfrac{2}{6}+\dfrac{1}{4}\) \(\Rightarrow\) \(\dfrac{13}{4}-\dfrac{11}{5}.\dfrac{10}{2}+\dfrac{3}{4}\le x< \dfrac{5}{14}.\dfrac{14}{15}+1\)
\(\dfrac{5}{3}.x=\dfrac{7}{12}\) \(\Rightarrow\dfrac{13}{4}-\dfrac{11}{5}.5+\dfrac{3}{4}\le x< \dfrac{5}{15}+1\)
\(x\) \(=\dfrac{7}{12}:\dfrac{5}{3}\) \(\Rightarrow\dfrac{13}{4}-11+\dfrac{3}{4}\le x< \dfrac{1}{3}+1\)
Vậy \(x=\dfrac{7}{30}\) \(\Rightarrow\dfrac{13}{4}+\dfrac{3}{4}-11\le x< \dfrac{4}{3}\)
\(\Rightarrow-7\le x< \dfrac{4}{3}\)
\(\Rightarrow\dfrac{-21}{3}\le\dfrac{3x}{3}< \dfrac{4}{3}\)
\(\Rightarrow-21\le3x< 4\)
Vậy \(x\in\left\{1;-1;-4;-5;-6;-7\right\}\)
Bài 2:
a) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
= \(\dfrac{1}{3}-\dfrac{1}{39}\)
= \(\dfrac{4}{13}\)
b) \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{32.35}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\)
= \(\dfrac{1}{2}-\dfrac{1}{35}\)
= \(\dfrac{33}{70}\)
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