mà đừng đổi ra số tụ nhiên nhế

\(2^7.\frac{3^4}{3^3}.2^5=2^{7+5}.3^{4-3}=2^{12}.3\)

\(\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3^1}{1}=\frac{4.3}{1}=\frac{12}{1}=12\)

Đặt : A = 3 + 3² + 3³ + ..... + 3¹⁰⁰

=> 3A = 3² + 3³ + 3⁴ + ..... + 3¹⁰¹

=> 3A - A = 3¹⁰¹ - 3

=> 2A = 3¹⁰¹ - 3

=> A = \(\frac{3^{101}-3}{2}\)

cảm ơn bạn nha at the speed of light thank you

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

2A=3A-A=(3+3^2+3^3+....+3^1000-(1+3+3^2+....+3^99) = 3^100-1

=>2A+1 = 3^100 = (3^5)^20 = 243^20

Vậy 2A+1 = 243^20

k mk nha

dư 0 nhé

a, \(\frac{6^5\cdot27^2}{7^3\cdot9^5}=\frac{2^5\cdot3^5\cdot\left(3^3\right)^2}{7^3\cdot\left(3^2\right)^5}=\frac{2^5\cdot3^5\cdot3^6}{7^3\cdot3^{10}}=\frac{2^5\cdot3^{11}}{7^3\cdot3^{10}}=\frac{2^5\cdot3}{7^3}\)

b, \(\frac{12^7\cdot9^3}{8^5\cdot27^3}=\frac{3^7\cdot2^{12}\cdot3^6}{2^{15}\cdot3^9}=\frac{2^{12}\cdot3^{13}}{2^{15}\cdot3^9}=\frac{3^4}{2^3}\)

c, \(\frac{20^6\cdot8^2}{16^3\cdot25^3}=\frac{2^{12}\cdot5^6\cdot2^6}{2^{12}\cdot5^6}=2^6\)

\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)

\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)

\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)

\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)

\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)

\(g\left(x\right)=-2x^4+4x^3+x-5\)

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`