3A-A= 3^2+3^3+....+3^101-3 -3^2-3^3-....-3^100

A=  (3^101-3 ) :2

                     A = 3 + 3² + 3³ + .... + 3¹⁰⁰

                   3A = 3² + 3³ + 3⁴ + ... + 3¹⁰¹

             3A - A = (3² + 3³ + 3⁴ + ... + 3¹⁰¹) - (3 + 3² + 3³ + ... + 3¹⁰⁰)

                  2A = 3¹⁰¹ - 3

              => A = \(\frac{3^{101}-3}{2}\)

              Ủng hộ mk nha !!! ^_^

:)?

giúp e với em cần gấp ạ

\(A=3^2\cdot4^3-3^2+333\)

\(=3^2\cdot63+333\)

\(=3^2\cdot\left(63+37\right)\)

\(=30^2\)

\(B=5\cdot3^2+4\cdot3^2=3^2\cdot9=9^2\)

d: Ta có: \(D=5^3+6^3+59\)

\(=125+216+59\)

\(=400=20^2\)

Ông ơi

thằng  trung anh ơiiiiiiiiiiiii

\(=3^{2+3}+2^{3+2}\)

\(=3^5+2^5\)

\(=243+32=275\)

\(3^2.3^3+2^3.2^2=3^{2+3}+2^{3+2}=3^5+2^5=243+32=275\)

a) \(2^3.3^2=8.9=72\)

b) \(5^{10}:5^7=5^2=25\)

c) \(2^6:2=2^5=32\)

d) \(7^4:7^4=7^0=1\)

e) \(9^5:9^5=9^0=1\)

a) 23.32=8.9=7223.32=8.9=72

b) 510:57=52=25510:57=52=25

c) 5

32

d) $1$

e) $1$

\(4A-3^{2023}\) hay \(4A=3^{2023}\) hả bạn

\(A=1-3+3^2-3^3+...+3^{2021}-3^{2022}\)

\(3A=3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\)

\(3A-A=\left(1-3+3^2-3^3+...+3^{2021}-3^{2022}\right)-\left(3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\right)\)

\(2A=3^{2023}-1\)

\(\Rightarrow A=\left(3^{2023}-1\right)\div2\)

\(\text{cái này mình sợ sai nên bạn có thể nhờ cô chữa}\)

4A - 3^{2023 nha bạn}

c: \(C=5\cdot4^3+2^4\cdot5+41\)

\(=5\cdot64+16\cdot5+41\)

\(=320+80+41\)

\(=441=21^2\)