\(A=\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\)

\(2A=1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\)

\(2A-A=\)\(\left(1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\right)-\)\(\left(\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\right)\)

\(A=1-\left(\frac{1}{2}\right)^{20}\)

gfvfvfvfvfvfvfv555

viết hẳn ra đi đừng viết mở ngoặc rồi đóng ngoặc

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)

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

⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\\\dfrac{2}{3}x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{2}{3}\\\dfrac{2}{3}x=0\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

3⁴⁰.8⁴¹- 8/4⁴⁰. 6^40-1=?

vầy hả

giúp mình nha 

Đặt A= \(\frac{1}{2}\)-\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)-\(\frac{1}{2^2}\)+....+\(\frac{1}{2^2}\)

=> 2A=1-\(\frac{1}{2}\)+\(\frac{1}{2^2}\)-\(\frac{1}{23}\)+...+\(\frac{1}{2^{98}}\)

=> 2A+A=1+\(\frac{1}{2^{99}}\)

=> 3A=1+\(\frac{1}{2^{99}}\)

=> A= \(\frac{1}{3}\)+\(\frac{1}{3.2^{99}}\)

Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)

=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)

\(Cm:\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Gọi biểu thức trên là A, ta có:

3A = 1-2/3+3/3^2-...-100/3^99

3A + A = [1-2/3+3/3^2-...-100/3^99] + [1/3-2/3^2+3/3^3-...-100/3^100]

4A = 1 - 1/3 + 1/3^2 - ... - 1/3^99 - 100/3^99 [1]

Gọi B = 1-1/3 + 1/3^2 - ... - 1/3^99

3B = 3 - 1 + 1/3 - 1/3^2 -...-1/3^2012

3B + B = [3-1+1/3-1/3^2-...-1/3^2012] + [1-1/3 + 1/3^2 - ... - 1/3^99]

4B = 3 - 1/3^99 

=> 4B < 3 => B < 1/4 [2]

Từ [1], [2] => 4A < B < 3/4 => A < 3/16 [đpcm]

MỎI TAY QUỚ

tk nha

Lúc đặt câu hỏi, bạn bấm vào góc trên cùng bên trái để gõ phép tính đẹp. Ý của bạn có phải là:

\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Bạn kia làm dúng rồi^_^

co ai giai dc ko cau 60 % .....