= 3⁹- 3⁹

= 0

# hok tốt !

=49-9+200=240

=49-9+200=240

A=1+3+3^2+3^3+3^4+3^5+3^6

3A=3+3^2+3^3+3^4+3^5+3^6+3^7

3A-A=(3+3^2+3^3+3^4+3^5+3^6+3^7)-(1+3+3^2+3^3+3^4+3^5+3^6)

A=3^7-1

Vì A =3^7-1 ; B =3^7-1

=> A=B

Sửa đề:

\(A=1+3+3^2+3^3+3^4+3^5+3^6\)

\(3A=3+3^2+...+3^7\)

\(3A-A=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)

\(2A=3^7-1\)

\(\Rightarrow A=\frac{3^7-1}{2}< 3^7-1=B\)

Vậy \(A< B\)

a)   \(x^3=216\)

=>  \(x^3=6^3\)

=>  \(x=6\)

ta có : 

S = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷

= (1 + 2) + (2² + 2³) + (2⁴ + 2⁵) + (2⁶ + 2⁷)

= (1 + 2) + 2²(1 + 2) + 2⁴(1 + 2) + 2⁶(1 + 2)

= (1 + 2)(1 + 2² + 2⁴ + 2⁶) 

= 3(1 + 2² + 2⁴ + 2⁶) \(⋮3\)(ĐPCM)

2S = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ 

S = (1+2 ) + (2² + 2³ ) + (2⁴ + 2⁵ ) + (2⁶ +2⁷)

S = 3 + 2²(1+2) + 2⁴(1+2) + 2⁶(1+2)

S = 3+2².3 + 2⁴.3 + 2⁶ .3 

S = 3(1+2² + 2⁴ + 2⁶ ) \(⋮\) 3

=> đpcm

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)(ĐPCM)

\((2x+3)^2=15^7:15^5\\\Rightarrow(2x+3)^2=15^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=15\\2x+3=-15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=12\\2x=-18\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-9\end{matrix}\right.\)

Vậy: ...