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Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
Các câu dễ tự làm nha:
\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)
a) \(1\dfrac{13}{15}.\left(-5\right)^2.3+\left(\dfrac{8}{15}-\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{28}{15}.25.3+\dfrac{13}{60}.\dfrac{24}{47}\)
\(=140+\dfrac{26}{235}=140\dfrac{26}{235}\)
b) \(\dfrac{\left(\dfrac{11^2}{200}+0,414:0,01\right)}{\dfrac{1}{12}-37.25+3\dfrac{1}{6}}\)
\(=\dfrac{\left(\dfrac{121}{200}-41,4\right)}{\dfrac{1}{12}-92519+\dfrac{19}{6}}\)
\(=\dfrac{2\dfrac{191}{207}}{-9251575}\)
a) \(\dfrac{-0.8:\left(\dfrac{4}{5}\cdot1.25\right)}{0.64-\dfrac{1}{5}}=\dfrac{\dfrac{-4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{5}}=\dfrac{\dfrac{-4}{5}:1}{\dfrac{16}{25}-\dfrac{5}{25}}=\dfrac{\dfrac{-4}{5}}{\dfrac{11}{25}}=\dfrac{-4}{5}\cdot\dfrac{25}{11}=\dfrac{-20}{11}\)
b) \(\left(13.71-1\dfrac{5}{6}\right)\cdot6-6\cdot13\cdot17=\left(\dfrac{1371}{100}-\dfrac{11}{6}\right)\cdot6-6\cdot13\cdot17=\dfrac{3563}{300}\cdot6-6\cdot13\cdot17=\dfrac{3563}{50}-6\cdot13\cdot17=\dfrac{3563}{50}-1326=\dfrac{-62737}{50}\)
c) \(\dfrac{\left(\dfrac{3}{5}+0.415+\dfrac{1}{200}\right):0.01}{30.75+\dfrac{1}{12}+3\dfrac{1}{6}}=\dfrac{\left(\dfrac{3}{5}+\dfrac{83}{200}+\dfrac{1}{200}\right):\dfrac{1}{100}}{\dfrac{123}{4}+\dfrac{1}{12}+\dfrac{19}{6}}=\dfrac{\dfrac{51}{50}:\dfrac{1}{100}}{34}=\dfrac{102}{34}=3\)
Bài 3 :
a) \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...........+\dfrac{1}{2017.2019}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+.........+\dfrac{2}{2017.2019}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2019}\)
\(\Leftrightarrow2A=\dfrac{672}{2019}\)
\(\Leftrightarrow A=\dfrac{336}{2019}\)
b) \(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+.........+\dfrac{1}{132}\)
\(\Leftrightarrow B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+............+\dfrac{1}{11.12}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+......+\dfrac{1}{11}-\dfrac{1}{12}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{12}=\dfrac{5}{12}\)
1.
Để \(\overline{25a89b}⋮2\Rightarrow b\in\left\{0;2;4;6;8\right\}\)
Để \(\overline{25a89b}\) chia 5 dư 3 \(\Rightarrow b\in\left\{3;8\right\}\)
Để thỏa mãn hai điều kiện trên thì \(b=8\)
Để \(\overline{25a898}⋮9\Rightarrow\left(2+5+a+8+9+8\right)⋮9\Leftrightarrow32+a⋮9\Rightarrow a=4\)
Vậy \(a=4;b=8\); số cần tìm là \(254898\)
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
\(.2.\)
\(a.\)
\(2x+\dfrac{1}{2}=-\dfrac{5}{3}\)
\(\Rightarrow2x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=-\dfrac{13}{6}:2=-\dfrac{13}{12}\)
Vậy : \(x=-\dfrac{13}{12}\)
\(b.\)
\(\dfrac{1}{7}-\dfrac{3}{5}x=\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{5}x=\dfrac{1}{7}-\dfrac{3}{5}=-\dfrac{16}{35}\)
\(\Rightarrow x=-\dfrac{16}{35}:\dfrac{3}{5}=-\dfrac{16}{21}\)
Vậy : \(x=-\dfrac{16}{21}\)
\(c.\)
\(\dfrac{3}{4}x+\dfrac{1}{2}=-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{4}x=-\dfrac{3}{5}-\dfrac{1}{2}=-\dfrac{11}{10}\)
\(\Rightarrow x=-\dfrac{11}{10}:\dfrac{3}{4}=-\dfrac{22}{15}\)
Vậy : \(x=-\dfrac{22}{15}\)
\(d.\)
\(-\dfrac{2}{15}-x=-\dfrac{3}{10}\)
\(\Rightarrow x=-\dfrac{2}{15}-\left(-\dfrac{3}{10}\right)=\dfrac{1}{6}\)
Vậy : \(x=\dfrac{1}{6}\)
=3/4*4/5*...*99/100
=3/100