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b,\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-2}{3}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-4}{6}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\).\(\frac{1}{6}\).\(\frac{3}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{18}\).\(\frac{3}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{6}\).\(\frac{1}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{12}\)
=\(\frac{8}{12}\)+\(\frac{1}{12}\)
=\(\frac{9}{12}\)=\(\frac{3}{4}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
a, \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)
\(=\frac{1}{3}.1-\frac{2}{3}\)
\(=\frac{1}{3}-\frac{2}{3}\)
\(=\frac{-1}{3}\)
b, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)
\(=\left(6-\frac{14}{5}\right).\frac{25}{8}+\frac{11}{8}.4\)
\(=\frac{16}{5}.\frac{25}{8}+\frac{11}{2}\)
\(=10+\frac{11}{2}\)
\(=\frac{31}{2}\)
1/3×(3/5+2/5)-2/15×1/5
1/3×1-2/15×1/5
1/3-2/15×1/5
1/3-2/75
25/75-2/75
23/75
(6-14/5)×25/8-11/8:4/1
16/5×25/8-11/8:4/1
10/1-11/8:4/1
10/1-11/8×1/4
10/1-11/32
320/32-11/32
309/32
(3+5+8+...+99)-(4+6+8+98)
Xét 3 + 5 + 8 + ... + 99 ( có 33 số hạng )
(3+5+8+...+99)-(4+6+8+98)
= [ ( 99 + 3 ) . 33 : 2 ] - 116
= 1683 - 116
= 1567
= \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
= \(\frac{2}{7}.2\)
= \(\frac{4}{7}\)
\(\frac{2}{7}.5\frac{1}{4}-\frac{2}{7}.3\frac{1}{4}\)
=> \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
=> \(\frac{2}{7}.2\)
=> \(\frac{4}{7}\)
#Hk_tốt
#Ngọc's_Ken'z
=49.(8+37+55)/{(2+98).[(98-2):2+1]}
=49.100/100.50
=49/50
\(\frac{49\times8+49\times37+49\times55}{2+4+6+8+....+96+98}.\)
= \(\frac{49\times\left(8+37+55\right)}{\left(98+2\right)\times\left[\left(98-2\right):2+1\right]:2}\)
= \(\frac{49\times100}{100\times49:2}\)
= \(\frac{49}{49:2}=\frac{49}{24,5}\)
= 2