1 phần 2+(1 phần 3 trừ 1 phần 6),1 phần 2 trừ 1 phần 3 trừ 1 phần 6,1phần 2 trừ 1 phần 6 + 1 phần 3
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viết rõ lại cái đề giùm tui tui mới làm được
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{6}{7}+\frac{2}{5}-\frac{1}{2}+2\frac{38}{91}-\frac{4}{5}=\frac{27300}{31850}+\frac{12740}{31850}-\frac{15925}{31850}+\) \(2\frac{13300}{31850}\)\(-\frac{25480}{31850}\)
\(=2\frac{27300+12740-15925+13300-25480}{31850}\)\(=2\frac{341}{910}\)
HỌC TỐT!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{4}.\dfrac{2}{3}-\dfrac{1}{6}.\dfrac{3}{4}-\dfrac{3}{4}.\dfrac{1}{2}\)
= \(\dfrac{3}{4}.\left(\dfrac{2}{3}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)
= \(\dfrac{3}{4}.0\)
= 0
\(\dfrac{3}{4}\cdot\dfrac{2}{3}-\dfrac{1}{6}\cdot\dfrac{3}{4}-\dfrac{3}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{3}{4}\left(\dfrac{2}{3}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{6}-\dfrac{1}{6}-\dfrac{3}{6}\right)\)
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{3}{2}-\frac{5}{6}:x=\frac{5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{3}{2}-\frac{5}{6}:x=\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{3}{2}-\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{41}{30}\)
\(\Leftrightarrow x=\frac{5}{6}:\frac{41}{30}\)
\(\Leftrightarrow x=\frac{25}{41}\)
b) \(x-\frac{6}{7}.\frac{14}{8}=\frac{1}{2}-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{3}{2}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{3}{2}\)
\(\Leftrightarrow x=\frac{8}{5}\)
c) \(x:\frac{6}{5}+\frac{2}{3}=\frac{7}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{7}{3}-\frac{2}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}.\frac{6}{5}\)
\(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(=-\frac{3.8...9999}{2^2.3^2...100^2}=-\frac{1.3.2.4...99.101}{2.2.3.3...100.100}=-\frac{\left(1.2....99\right).\left(3.4...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)
\(< -\frac{100}{200}=\frac{1}{2}=B\)
=> A < B