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`Answer:`
Bài 1:
a. \(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}-\frac{2}{3}x+\frac{1}{3}=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{6}-\frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow-\frac{2}{3}=\frac{2}{3}-\frac{5}{6}\)
\(\Leftrightarrow-\frac{2}{3}x=-\frac{1}{6}\)
\(\Leftrightarrow x=-\frac{1}{6}:-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b. \(\frac{3}{x+5}=15\%\left(ĐKXĐ:x\ne-5\right)\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow\frac{60}{20\left(x+5\right)}=\frac{3\left(x+5\right)}{20\left(x+5\right)}\)
\(\Leftrightarrow60x=3x+15\)
\(\Leftrightarrow-3x=-45\)
\(\Leftrightarrow x=15\)
Bài 2:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(a,\Rightarrow A=-1\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{9.10}\right)\)
\(\Rightarrow A=-1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(\Rightarrow A=-1\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)
\(\Rightarrow A=\dfrac{-3}{20}\)
Bài 2:
\(a,\dfrac{1717}{8585}=\dfrac{1717:1717}{8585:1717}=\dfrac{1}{5};\dfrac{1313}{5151}=\dfrac{1313:101}{5151:101}=\dfrac{13}{51}\\ \dfrac{1}{5}=\dfrac{51}{255}< \dfrac{65}{255}=\dfrac{13}{51}\\ \Rightarrow\dfrac{1717}{8585}< \dfrac{1313}{5151}\)
\(b,\dfrac{201201}{202202}=\dfrac{201201:1001}{202202:1001}=\dfrac{201}{202}=\dfrac{201\cdot1001001}{202\cdot1001001}=\dfrac{201201201}{202202202}\)
