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\(\dfrac{8}{13}+\dfrac{4}{9}+\dfrac{1}{3}+\dfrac{5}{13}+3\)
\(=1+\dfrac{7}{9}+3\)
\(=4+\dfrac{7}{9}=\dfrac{43}{9}\)
a/\(4,19\times50-132:\dfrac{2}{3}\)
\(=209,5-198\)
\(=11,5\)
b/\(637,38:18\times2,5\)
\(=35,41\times2,5\)
\(=88,525\)
c/\(56,32-13,4\times2,4\)
\(=56,32-32,16\)
\(=24,16\)
d/\(\dfrac{2}{5}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}:\dfrac{3}{10}\)
\(=\dfrac{4}{3}\)
\(1,\\ a,=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times1=\dfrac{7}{19}\\ b,=\dfrac{2}{5}+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{2}{5}+1=\dfrac{7}{5}\\ 2,\\ a,=15\times\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)\\ =15\times\left(\dfrac{2121:101}{4343:101}+\dfrac{222222:10101}{434343:10101}\right)\\ =15\times\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\times\dfrac{43}{43}=15\times1=15\)
\(3,\)
Cạnh \(AC=\) chu vi ABC \(-AB-BC=\dfrac{4}{5}-\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{3}{5}-\dfrac{1}{4}=\dfrac{7}{20}\left(m\right)\)
Vì \(\dfrac{7}{20}>\dfrac{5}{20}>\dfrac{4}{20}\Rightarrow\dfrac{7}{20}>\dfrac{1}{4}>\dfrac{1}{5}\) nên \(AC>BC>AB\)
a: 2h20p*2=4h40p
1h25p*3=3h75p=4h15p
b: 3h48p:3=1h16p
4h15p:5=255p:5=51p
c: 2h15p*2+6h15p:5
=4h30p+375p:5
=270p+75p=345p
d: \(=9h75p+288p:3\)
=615p+96p=711p
Có: \(B=1+4+4^2+...+4^{2009}\)
=> \(4.B=4.\left(1+4+4^2+...+4^{2019}\right)\)
\(4B=4+4^2+4^3+...+4^{2020}\)
=> \(4B-B=\left(4+4^2+4^3+...+4^{2020}\right)-\left(1+4+4^2+...+4^{2019}\right)\)
\(3B=\left(4-4\right)+\left(4^2-4^2\right)+...+\left(4^{2019}-4^{2019}\right)+\left(4^{2020}-1\right)\)
\(3B=4^{2020}-1\)
=> \(3B+1=4^{2020}-1+1\)
\(3B+1=4^{2020}\)
Vậy 3B + 1 là lũy thừa của 4.
\(B=1+4+4^2+......+4^{2019}\)
\(\Rightarrow4B=4+4^2+4^3+.......+4^{2020}\)
\(\Rightarrow4B-B=3B=4^{2020}-1\)
Ta có: \(3B+1=4^{2020}-1+1=4^{2020}\)là lũy thừa của 4 ( đpcm )
\(1\frac{1}{2}+1\frac{1}{3}=\frac{11}{6}\)
\(3\frac{1}{2}\div2\frac{1}{4}=\frac{14}{9}\)
Học tốt
a) \(1\frac{1}{2}+1\frac{1}{3}\)
\(=\frac{3}{2}+\frac{4}{3}\)
\(=\frac{17}{6}\)
b) \(3\frac{1}{2}:2\frac{1}{4}\)
\(=\frac{7}{2}:\frac{9}{4}\)
\(=\frac{14}{9}\)