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\(\frac{3}{6.8}+\frac{3}{8.10}+.......+\frac{3}{198.200}\)
\(=\frac{3}{2}.\left(\frac{2}{6.8}+\frac{2}{8.10}+........+\frac{2}{198.200}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+........+\frac{1}{198}-\frac{1}{200}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{6}-\frac{1}{200}\right)\)
\(=\frac{3}{2}.\frac{97}{600}=\frac{97}{400}\)
\(3.\left(\frac{2}{6.8}+\frac{2}{8.10}+....+\frac{2}{198.200}\right).\frac{1}{2}\)
=\(3.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{198}{200}\right).\frac{1}{2}\)
=\(3.\left(\frac{1}{6}-\frac{1}{200}\right).\frac{1}{2}\)
=.\(3.\frac{97}{600}.\frac{1}{2}\)=97/400
\(\frac{2}{4}+\frac{1}{4}=\frac{3}{4}\)\(\\ \frac{1}{3}+\frac{2}{3}=1\)\(\frac{9}{10}+\frac{1}{10}=1\)
ta có \(\frac{3}{7}=\frac{3\times3}{7\times3}=\frac{9}{21}\)(quy đồng tử)
So sánh \(\frac{9}{21}\)và \(\frac{9}{17}\)ta có:
\(21>17\Rightarrow\frac{9}{17}>\frac{9}{21}\Rightarrow\frac{9}{17}>\frac{3}{7}\)
VẬY: \(\frac{9}{17}>\frac{3}{7}\)
Ta có:
11/32 > 11/33 = 1/3 = 9/27 > 9/28
=> 11/32 > 9/28
Vậy 9/28 < 11/32
\(\frac{4}{15}:\frac{4}{7}< x< \frac{2}{5}.\frac{10}{3}\Leftrightarrow\frac{7}{15}< x< \frac{20}{15}\)
\(\Rightarrow x\in\left(8;9;10;11;12;13;14;15;16;17;18;19\right)\)
Ta có : \(\frac{4}{15}:\frac{4}{7}< x< \frac{2}{5}\cdot\frac{10}{3}\)
\(\Rightarrow\frac{4}{15}\cdot\frac{7}{4}< x< \frac{2}{5}\cdot\frac{10}{3}\)
\(\Rightarrow\frac{7}{15}< x< \frac{20}{15}\)
\(\Rightarrow7< x< 20\)
\(\Rightarrow x\in\left\{8;9;10;11;12;...;19\right\}\)
TL
\(\frac{10}{24};\frac{9}{8_{\left(3\right)}}\left(MSC:24\right)\)
\(\frac{9}{8}=\frac{9x3}{8x3}=\frac{27}{24}\), giữ nguyên phân số \(\frac{10}{24}\)
Vậy quy đồng mẫu số hai phân số \(\frac{10}{24},\frac{9}{8}\) ta được \(\frac{27}{24},\frac{10}{24}\)
_HT_
40:\(\frac{3}{10}\)=\(\frac{400}{3}\)nha bn
\(40:\frac{3}{10}\)
\(=40:0,3\)
\(=133,3333333=\frac{400}{3}\)