4\9 +1\8 +5\9 + 7\8 b) 1\3 + 2\3 +5\12 + 4\3 + 1\12
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\(\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{8}{9}\right)=1+1+1+1=4\)
A = (2 + 22) + (23 + 24 ) +…(2199 + 2200)
A = 6 + 22 (2 + 22 ) +… + 2198 (2 + 22)
A = 6 + 22 (6 ) +… + 2198 (6)
A = 6(1 + 22 +… + 2198)
Vậy A chia hết cho 6
a) \(\left(\frac{2}{7}+\frac{3}{8}+\frac{12}{7}+\frac{5}{8}\right).\frac{19}{36}\)
\(=\left\{\left(\frac{2}{7}+\frac{12}{7}\right)+\left(\frac{3}{8}+\frac{5}{8}\right)\right\}.\frac{19}{36}\)
\(=\left(\frac{14}{7}+\frac{8}{8}\right).\frac{19}{36}\)
\(=\left(2+1\right).\frac{19}{36}\)
\(3.\frac{19}{36}=\frac{19}{12}\)
\(a.60\times\left(\frac{7}{12}+\frac{4}{15}\right)\)
\(=60\times\left(\frac{35}{60}+\frac{16}{60}\right)\)
\(=60\times\frac{51}{60}\)
\(=51\)
\(b.\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\)
\(=\frac{1\times2\times3\times4\times5\times6\times7\times8}{2\times3\times4\times5\times6\times7\times8\times9}\)
\(=\frac{1}{9}\)
\(\frac{9}{10}\)+\(\frac{7}{9}\)+\(\frac{5}{8}\)+\(\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
\(=\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)\)\(+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=1+1+1+1+1\)
\(=5\)
9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
= ( 9/10 + 1/10 ) + ( 7/9 + 2/9 ) + ( 5/8 + 3/8 ) + ( 3/7 + 4/7 ) + ( 3/5 + 2/5 )
= 1 + 1 + 1 + 1 + 1
= 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)
= \(1+1+1+1+1\)
= \(1\times5\)
= \(5\)
Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :
A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)
A = 1 + 1 + 1 + 1 + 1
A = 5
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)\(=\frac{9}{10}+\frac{1}{10}+\frac{7}{9}+\frac{2}{9}+\frac{5}{8}+\frac{3}{8}+\frac{3}{7}+\frac{4}{7}+\frac{3}{5}+\frac{2}{5}\)
\(=\left[\frac{9}{10}+\frac{1}{10}\right]+\left[\frac{7}{9}+\frac{2}{9}\right]+\left[\frac{5}{8}+\frac{3}{8}\right]+\left[\frac{3}{7}+\frac{4}{7}\right]+\left[\frac{3}{5}+\frac{2}{5}\right]\)
\(=1+1+1+1+1\)
\(=5\)
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= 1 + 1 + 1 + 1 + 1
= 5
1+2+3+...+9
=(1+9)+(2+8)+...+5
=10+10+...+5
=10.4+5
=45
HT
\(a,\\ =\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\left(\dfrac{1}{8}+\dfrac{7}{8}\right)=1+1=2\\ b,\\ =\left(\dfrac{1}{3}+\dfrac{4}{3}+\dfrac{2}{3}\right)\left(\dfrac{5}{12}+\dfrac{1}{12}\right)\\ =\dfrac{7\times2}{3\times2}+\dfrac{1\times3}{2\times3}=\dfrac{14+3}{6}=\dfrac{17}{6}\)
\(a,\dfrac{4}{9}+\dfrac{1}{8}+\dfrac{5}{9}+\dfrac{7}{8}=\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\left(\dfrac{1}{8}+\dfrac{7}{8}\right)=\dfrac{9}{9}+\dfrac{8}{8}=1+1=2\\ b,\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{5}{12}+\dfrac{4}{3}+\dfrac{1}{12}=\left(\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{4}{3}\right)+\left(\dfrac{5}{12}+\dfrac{1}{12}\right)=\dfrac{7}{3}+\dfrac{6}{12}=\dfrac{7}{3}+\dfrac{1}{2}=\dfrac{14}{6}+\dfrac{3}{6}=\dfrac{17}{6}\)