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a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)
\(\Rightarrow n+1=50\)
\(\Rightarrow n=49\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)
\(\Rightarrow2n+1=51\)
\(\Rightarrow2n=50\)
\(\Rightarrow n=25\)
a. \(\frac{1}{2}\) - ( \(\frac{1}{3}\) + \(\frac{1}{4}\) ) < x < \(\frac{1}{48}\) - ( \(\frac{1}{16}\) - \(\frac{1}{6}\) )
\(\frac{1}{2}\) - \(\frac{7}{12}\) < x < \(\frac{1}{48}\) - \(\frac{-5}{48}\)
\(\frac{-1}{12}\) < x < \(\frac{1}{8}\)
Đề bài yêu cầu tìm x thuộc tập hợp gì bạn ơi. Bạn viết thiếu rồi .
a) \(\frac{5}{9}:\left(\frac{5}{12}-\frac{1}{11}\right)-\frac{5}{9}:\left(\frac{-1}{5}-\frac{2}{3}\right)\)
= \(\frac{5}{9}:\left(\frac{55}{132}-\frac{12}{132}\right)-\frac{5}{9}:\left(\frac{-3}{15}-\frac{10}{15}\right)\)
= \(\frac{5}{9}:\frac{43}{132}-\frac{5}{9}:\frac{-13}{15}\)
= \(\frac{5}{9}\times\frac{132}{43}-\frac{5}{9}\times\frac{-15}{13}\)
=\(\frac{5}{9}\times\left(\frac{132}{43}-\frac{-15}{13}\right)\)
=\(\frac{5}{9}\times\frac{2361}{559}\)( Đến đây bạn tự quy đồng mẫu nha)
=\(\frac{3935}{1677}\)
a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)
\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)
\(=\frac{12}{15}\)
\(=\frac{4}{5}\)
c, \(\frac{3}{8}.3\frac{1}{3}\)
\(=\frac{3}{8}.\frac{10}{3}\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)
\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)
\(=\frac{-3}{5}+\frac{-60}{10}\)
\(=\frac{-3}{5}+\frac{-30}{5}\)
\(=\frac{-33}{5}\)
e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)
\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)
\(=\frac{2}{5}.10\)
\(=4\)
f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.-14\)
\(=-6\)
~Study well~
#KSJ
Bạn tham khảo nhé
a ) Ta có :
\(\left(-\frac{1}{5}\right)^{300}=\left(\frac{1}{5}\right)^{300}=\frac{1}{5^{300}}=\frac{1}{\left(5^3\right)^{100}}=\frac{1}{125^{100}}\)
\(\left(-\frac{1}{3}\right)^{500}=\left(\frac{1}{3}\right)^{500}=\frac{1}{3^{500}}=\frac{1}{\left(3^5\right)^{100}}=\frac{1}{243^{100}}\)
Do \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\left(125^{100}< 243^{100}\right)\)
\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)
b )
Ta có :
\(2550^{10}=\left(50.51\right)^{10}=50^{10}.51^{10}\)
\(50^{20}=50^{10}.50^{10}\)
Do \(50^{10}.51^{10}>50^{10}.50^{10}\)
\(\Rightarrow50^{20}< 2550^{10}\)
c )
Ta có :
\(2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(5^{50}=\left(5^2\right)^{25}=25^{25}\)
Do \(16^{25}< 25^{25}< 27^{25}\)
\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{303}=\frac{49}{303}\)
\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{2550}\)
\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{50\cdot51}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{50}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}\)
\(=\frac{16}{51}\)