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bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\)
\(x\times\dfrac{11}{16}=2\)
\(x=2:\dfrac{11}{16}\)
\(x=\dfrac{32}{11}\)
Bài 1 :
\(\dfrac{x}{16}\times\left(2017-1\right)=2\)
\(\dfrac{x}{16}\times2016=2\)
\(\dfrac{x}{16}=\dfrac{2}{2016}\)
\(x=\dfrac{2}{2016}\times16\)
\(x=\dfrac{1}{63}\)
1a,\(\frac{8}{5}\)=1\(\frac{3}{5}\)b,\(\frac{9}{2}\)=4\(\frac{1}{2}\)c,\(\frac{7}{4}\)=1\(\frac{3}{4}\)d,\(\frac{10}{3}\)
2a,1\(\frac{3}{4}\)+1\(\frac{5}{8}\)=\(\frac{7}{4}\)+\(\frac{13}{8}\)=\(\frac{14}{8}\)+\(\frac{13}{8}\)=\(\frac{27}{8}\)b,3\(\frac{4}{5}\)x2\(\frac{6}{8}\)=\(\frac{19}{5}\)x\(\frac{22}{8}\)=\(\frac{418}{40}\)=\(\frac{209}{20}\)
3,3\(\frac{2}{5}\)=\(\frac{17}{5}\),3\(\frac{4}{5}\)=\(\frac{19}{5}\) , 8\(\frac{2}{4}\)=\(\frac{34}{4}\),10\(\frac{6}{7}\)=\(\frac{76}{7}\) , 5\(\frac{1}{2}\)=\(\frac{11}{2}\),5\(\frac{3}{4}\)=\(\frac{23}{4}\)
=>\(\frac{17}{5}\)<\(\frac{19}{5}\)=>3\(\frac{2}{5}\)<3\(\frac{4}{5}\) =>\(\frac{34}{4}\)<\(\frac{76}{7}\)=>8\(\frac{2}{4}\)<10\(\frac{6}{7}\) =>\(\frac{11}{2}\)<\(\frac{23}{4}\)=>5\(\frac{1}{2}\)<5\(\frac{3}{4}\)
k mik nha ......
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Ta có : \(9+\frac{9}{2}+\frac{9}{4}+......+\frac{9}{128}+\frac{9}{256}\)
\(=9\left(1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}+\frac{1}{256}\right)\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}+\frac{1}{256}\)
=> \(2A=2+1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)
=> \(2A-A=2-\frac{1}{256}\)
=> \(A=\frac{511}{256}\)
Thay A vào ta có : \(9.\frac{511}{256}=\frac{4599}{256}\)
1a)\(\frac{5}{3}\)=\(\frac{5x4}{3x4}\)=\(\frac{20}{12}\); \(\frac{1}{4}\)=\(\frac{1x3}{4x3}\)=\(\frac{3}{12}\)
b)\(\frac{3}{8}\)=\(\frac{3x3}{8x3}\)=\(\frac{9}{24}\); \(\frac{7}{24}\)
c)\(\frac{1}{2}\)=\(\frac{1x15}{2x15}\)=\(\frac{15}{30}\); \(\frac{2}{3}\)=\(\frac{2x10}{3x10}\)=\(\frac{20}{30}\); \(\frac{3}{5}\)=\(\frac{3x6}{5x6}\)=\(\frac{18}{30}\)
2a)\(\frac{11}{8}\)>\(\frac{11}{9}\)
b)\(\frac{4}{9}\)<\(\frac{3}{5}\)
c)\(\frac{6}{5}\)>\(\frac{5}{6}\)
Bài làm:
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}.y=\frac{4}{3}\)
\(\Leftrightarrow y=\frac{4}{3}:\frac{10}{11}=\frac{4}{3}.\frac{11}{10}=\frac{22}{15}\)
Chú ý dấu \(\left(.\right)\)là dấu \(\left(\times\right)\)
Vậy \(y=\frac{22}{15}\)
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{10}{11}.y=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{11}.y=\frac{2}{3}\)
\(\Rightarrow y=\frac{2}{3}:\frac{5}{11}=\frac{22}{15}\)
LƯU Ý:các dấu chấm(.) là dấu nhân ^^.