A) \(\dfrac{5}{10}-\dfrac{2}{15}\)

\(=\dfrac{1}{2}-\dfrac{2}{15}\)

\(=\dfrac{15}{30}-\dfrac{4}{30}\)

\(=\dfrac{11}{30}\)

B)\(\dfrac{5}{20}-\dfrac{1}{6}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}\)

\(=\dfrac{6}{24}-\dfrac{4}{24}\)

\(=\dfrac{2}{24}=\dfrac{1}{12}\)

C) \(\dfrac{6}{18}-\dfrac{6}{24}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}\)

\(=\dfrac{4}{12}-\dfrac{3}{12}\)

\(=\dfrac{1}{12}\)

D) \(\dfrac{5}{9}-\dfrac{3}{12}\)

\(=\dfrac{5}{9}-\dfrac{1}{4}\)

\(=\dfrac{20}{36}-\dfrac{9}{36}\)

\(=\dfrac{11}{36}\).

a: =1/2-2/15

=15/30-4/30

=11/30

b: =1/4-1/6

=3/12-2/12

=1/12

c: =1/3-1/4

=1/12

d: =20/36-9/36

=11/36

a) \(\dfrac{5}{6}-\dfrac{2}{14}\)

\(=\dfrac{5}{6}-\dfrac{1}{7}\)

\(=\dfrac{35}{42}-\dfrac{6}{42}\)

\(=\dfrac{29}{42}\)

b) \(\dfrac{5}{20}-\dfrac{1}{6}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}\)

\(=\dfrac{3}{12}-\dfrac{2}{12}\)

\(=\dfrac{1}{12}\)

c) \(\dfrac{5}{9}-\dfrac{3}{12}\)

\(=\dfrac{5}{9}-\dfrac{1}{4}\)

\(=\dfrac{20}{36}-\dfrac{9}{36}\)

\(=\dfrac{11}{36}\)

d) \(8-\dfrac{4}{6}\)

\(=\dfrac{48}{6}-\dfrac{4}{6}\)

\(=\dfrac{44}{6}=\dfrac{22}{3}\).

cái đó là toán lớp 5 nha bẩm lộn ấ

a, 5/10 - 2/15 = 1/2 - 2/15 = 15/30 - 4/30 = 11/30

b,5/20 - 1/6 = 1/4 - 1/6 = 3/12 - 2/12 = 1/12

c,6/18 - 6/24 = 1/3 - 1/4 = 4/12 - 3/12 = 1/12 

d,5/9 - 3/12 = 5/9 - 1/4 = 20/36 - 9/36 = 11/36 
chắc là đúng á , tại lớp 8 rồi ko nhớ lắm , mà vote mình nhá

ròi cái d lòi đâu ra zị

a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)

Ta có A=1/5+1/20+1/44+...+1/170

=>A=1/10+1/40+1/88+...+1/340

   A=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+......+\(\frac{1}{17.20}\)

   A=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+.....+1/17-1/20)

   A=1/3.(1/2-1/20)

   A=\(\frac{3}{20}\)

`(5sqrt{1/5}+1/2sqrt{20}-5/4sqrt{4/5}+sqrt{5}):2/5

`=(sqrt5+1/2*2sqrt5-sqrt{5/4}+sqrt5):2/5`

`=(sqrt5+sqrt5+sqrt5-sqrt5/2):2/5`

`=(5/2*sqrt5):2/5`

`=25/4sqrt5`

`1/3sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1 1/3}`

`=1/3*4sqrt3+3*5sqrt3-3sqrt3-10sqrt{4/3}`

`=4/sqrt3+15sqrt3-3sqrt3-20/sqrt3`

`=12sqrt3-16/sqrt3`

1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

a,

B = 1 + 5 + 5^2 + 5^3 + ... + 5^100

5B = 5 + 5^2 + 5^3 + ... + 5^101

5B - B = [5 + 5^2 + 5^3 + ... + 5^101] - [1 + 5 + 5^2 + 5^3 + ... + 5^100]

4B = 5 + 5^2 + 5^3 + ... + 5^101 - 1 - 5 - 5^2 - 5^3 - ... - 5^100

4B = 5^101 - 1

B = \(\frac{5^{101}-1}{4}\)

b,

A = 1 - 3 + 3^2 - 3^3 + ... + 3^20 - 3^21

3A = 3 - 3^2 + 3^3 - 3^4 + ... + 3^21 - 3^22

3A - A = [3 - 3^2 + 3^3 - 3^4 + ... + 3^21 - 3^22] - [1 - 3 + 3^2 - 3^3 + ... + 3^20 - 3^21]

2A = 3 - 3^2 + 3^3 - 3^4 + ... + 3^21 - 3^22 - 1 + 3 - 3^2 + 3^3 + ... - 3^20 + 3^21

2A = 2[3 + 3^3 + 3^5 + ... + 3^21] - 2[3^2 + 3^4 + ... + 3^20] - 1

Đặt C = 3 + 3^3 + 3^5 + ... + 3^21 

=> 3^2C = 3^3 + 3^5 + 3^7 + ... + 3^23

=> 9C - C = [3^3 + 3^5 + 3^7 + ... + 3^23] - [3 + 3^3 + 3^5 + ... + 3^21]

=> 8C = 3^3 + 3^5 + 3^7 + ... + 3^23 - 3 - 3^3 - 3^5 - ... - 3^21

=> 8C = 3^23 - 3

=> C = 3^23 - 3 / 8

=> 2[3 + 3^3 + 3^5 + ... + 3^21] = 3^23 - 3 / 8 * 2 = 3^23 - 3 / 4

Đặt D = 3^2 + 3^4 + ... + 3^20 

=> 3^2D = 3^4 + 3^6 + ... + 3^22

=> 9D - D = [3^4 + 3^6 + ... + 3^22] - [3^2 + 3^4 + ... + 3^20]

=> 8D = 3^4 + 3^6 + ... + 3^22 - 3^2 - 3^4 - ... - 3^20

=> 8D = 3^22 - 9

=> D = 3^22 - 9 / 8

=> 2[3^2 + 3^4 + ... + 3^20] = 3^22 - 9 / 8 * 2 = 3^22 - 9 / 4

=> A = 3^23 - 3 / 4 - 3^22 - 9 / 4 - 1

\(\Rightarrow A=\frac{3^{23}-3-3^{22}-9}{4}-1=\frac{3^{22}\left[3-1\right]-12}{4}=\frac{3^{22}\cdot2-12}{4}\)

\(=\frac{6\left[3^{21}-2\right]}{4}=\frac{3\left[3^{21}-2\right]}{2}=5230176601\)

Mình chỉ biết làm thế thôi, sai thì mong mn sửa lại giúp nhé