10/21

\(=\dfrac{91}{105}-\dfrac{101}{105}+\dfrac{4}{7}=-\dfrac{2}{21}+\dfrac{12}{21}=\dfrac{10}{21}\)

10/21

  13/15 + 4/7 - 101/105

= 91/105 + 60/105 - 101/105

= 50/105 = 10/21

học tốt

13/5+4/7-101/105

=13.21/5.21+ 4.15/7.15-101/105

=273/105+50/105 -101/105

=222/105

232/105

\(\dfrac{13}{5}+\dfrac{4}{7}-\dfrac{101}{105}=\dfrac{273}{105}+\dfrac{60}{105}-\dfrac{101}{105}=\dfrac{232}{105}\)

Xét chữ số tận cùng của 7¹⁰¹

7¹⁰¹ = 7¹⁰⁰.7 = (7⁴)²⁵.7 = (....1)²⁵.7

= ....1.7 = ....7

=> 7¹⁰¹ chia 10 dư 7

\(a,12-\left(4+105-13\right)+\left(4+105-3\right)\)

\(=12-4-105+13+4+105-3\)

\(=12+\left(-4+4\right)+\left(-105+105\right)+\left(13-3\right)\)

\(=12+0+0+10\)

\(=22\)

\(b,-15+\left(-17+21-35\right)+\left(4+105-13\right)\)

\(=-15-17+21-35+4+105-13\)

\(=\left(-15-35\right)+\left(-17-13\right)+\left(21+4+105\right)\)

\(=-50-30+130\)

\(=-50+100\)

\(=50\)

\(c,\left(\left|-5\right|+\left|+7\right|+\left|-9\right|\right)-\left(\left|+5\right|+\left|-7\right|+9\right)\)

\(=\left(5+7+9\right)-\left(5+7+9\right)\)

\(=0\)

a: =91/105+60/105-101/105

=50/105=10/21

c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)

d: =2-2/9

=18/9-2/9

=16/9

e: =24/36-9/36+8/36

=23/36

g: =5/2+1/2

=3

A=1/2-1/3+1/3-1/4+...+1/149-1/150

=1/2-1/150

=37/75

B=3A=37/25

\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{149\cdot150}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{149}-\frac{1}{150}\)

\(A=\frac{1}{2}-\frac{1}{150}\)

\(A=\frac{75}{150}-\frac{1}{150}\)

\(A=\frac{74}{150}=\frac{37}{75}\)

Vậy \(A=\frac{37}{75}\)

a) <     b) >     c) <