nhóm 1: \(\dfrac{5}{3}x^2y;\dfrac{-1}{3}x^2y;x^2y;\dfrac{-2}{5}x^2y\)

nhóm 2: \(xy^2;-2xy^2;\dfrac{1}{4}xy^2\)

nhóm 3: xy

2.\(25xy^2+55xy^2+75xy^2=155xy^2\)

3. thay x=1 và y=-1 vào biểu thức ta đc:

\(\dfrac{1}{2}.1^5.\left(-1\right)-\dfrac{3}{4}1^5.\left(-1\right)1^5.\left(-1\right)=\dfrac{1}{4}\)

\(B=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)

\(\Rightarrow2B-B=2^{101}-2\)

bài 1

2¹⁰¹ - 2

\(A=1+3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)\left(1+3^4+...+9^{96}\right)\)

\(\Rightarrow A=40\left(1+3^4+...+9^{96}\right)⋮40\)

b, -418 - {- 418 - [ -418 - (-418) + 2021]}

= -481 - { -418  - [ 0 + 2021]}

= -481 + 418 + 2021

= 2021 

d, 23 - 501 - 343 + 61 - 257 + 16 - 499

=  (23 + 61 + 16) - (501 + 499) - (343 + 257)

= 100 - 1000 - 600

= 100 - 1600

= -1500 

e, 743 - 231 + (-495) - (-69) - 38 + (-117)

= 512 - 426 - 155

= 86 - 155

= - 69 

a) \(S=1+2+2^2+2^3+...+2^{2022}=\dfrac{2^{2022+1}-1}{2-1}=2^{2023}-1\)

b) \(S=1+4+4^2+4^3+...+4^{2022}=\dfrac{4^{2022+1}-1}{4-1}=\dfrac{4^{2023}-1}{3}\)

\(S=1+2+2^2+2^3+...+2^{2022}\\ 2S=2+2^2+2^3+2^4+...+2^{2023}\\ 2S-S=2+2^2+2^3+2^4+...+2^{2023}-1-2-2^2-2^3-...-2^{2022}\\ S=2^{2023}-1\\ S=4+4^2+4^3+...+4^{2022}\\ 4S=4^2+4^3+4^4+...+4^{2023}\\ 4S-S=4^2+4^3+4^4+...+4^{2023}-4-4^2-4^3-...-4^{2023}\\ 3S=4^{2023}-4\\ S=\dfrac{4^{2023}-4}{3}\)

a)Ta có:  \(5^{36}=5^{3.12}=\left(5^3\right)^{12}=125^{12}\)

              \(11^{24}=11^{2.12}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125>121\Rightarrow125^{12}>121^{12}\)

\(\Rightarrow5^{36}>11^{24}\)

b) Ta có: \(625^5=\left(5^4\right)^5=5^{20}\)

              \(125^7=\left(5^3\right)^7=5^{21}\)

Vì \(20< 21\Rightarrow5^{20}< 5^{21}\)

\(\Rightarrow625^5< 125^7\)

c) Ta có: \(3^{2n}=\left(3^2\right)^n=9^n\)

                \(2^{3n}=\left(2^3\right)^n=8^n\)

Vì \(9>8\Rightarrow9^n>8^n\)( do \(n>0\))

\(\Rightarrow3^{2n}>2^{3n}\)

d)Ta có:  \(5^{23}=5.5^{22}< 6.5^{22}\)

\(\Rightarrow5^{23}< 6.5^{22}\)

a. 5^36=(5^3)^12

               =125^12

11^24=(11^2)^12

         = 121^12

Vì 125^12>121^12 nên 5^36>11^24

b. Ta có: 625^5 =(5^4)^5

                         = 5^20

               125^7=(5^3)^7

                        = 5^21

 Vì 5^20<5^21 nên 625^5<125^7

\(=\dfrac{20}{21}x\dfrac{21}{22}x\dfrac{22}{23}x...x\dfrac{1999}{2000}\)

\(=\dfrac{20}{2000}=\dfrac{1}{100}\)

=20/21x21/22x22/23x..............x1998/1999x1999/2000

=20x21x22x23x.....................x1998x1999/21x22x23x24x...............x1999x2000

=20/2000

1/100

A = (1- \(\dfrac{1}{21}\))\(\times\)( 1 - \(\dfrac{1}{22}\))\(\times\)(1-\(\dfrac{1}{23}\))\(\times\)...\(\times\)(1-\(\dfrac{1}{1999}\))\(\times\)(1- \(\dfrac{1}{2000}\))

A = \(\dfrac{21-1}{21}\)\(\times\)\(\dfrac{22-1}{22}\)\(\times\)\(\dfrac{22-1}{23}\)\(\times\)...\(\times\)\(\dfrac{1999-1}{1999}\)\(\times\)\(\dfrac{2000-1}{2000}\)

A = \(\dfrac{20}{21}\)\(\times\)\(\dfrac{21}{22}\)\(\times\)\(\dfrac{1998}{1999}\)\(\times\)\(\dfrac{1999}{2000}\)

A = \(\dfrac{21\times....\times1999}{21\times...\times1999}\) \(\times\) \(\dfrac{20}{2000}\)

A= \(\dfrac{1}{100}\)

\(A=2+2^2+...+2^{20}\)

\(2A=2^2+2^3+...+2^{21}\)

\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)

\(A=2^{21}-2\)

___________

\(B=5+5^2+...+5^{50}\)

\(5B=5^2+5^3+...+5^{51}\)

\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)

\(4B=5^{51}-5\)

\(B=\dfrac{5^{51}-5}{4}\)

___________

\(C=1+3+3^2+...+3^{100}\)

\(3C=3+3^2+...+3^{101}\)

\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)

\(2C=3^{101}-1\)

\(C=\dfrac{3^{101}-1}{2}\)

2A= 2(2+2²+2³+...+2¹⁹+2²⁰)

2A= 2²+2³+2⁴+...+2²⁰+2²¹

2A-A=(2²+2³+2⁴+...+2²⁰+2²¹)-(2+2²+2³+...+2¹⁹+2²⁰)

A=2²¹-2

Vậy A=2²¹-2

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