gọi biểu thức 2^4+2^8+....+2^2016 là A ta có

A=2^4+2^8+.....+2^2016

8A=2^4+2^8+.....+2^2010

8A-A=2-2^2010

7A=1+2-2^2010

chưa hiểu

A = 1 + 2⁴ + 2⁸ + ...... + 2²⁰¹² + 2²⁰¹⁶

2⁴A = 2⁴ + 2⁸ + 2¹² + ..... + 2²⁰¹⁶ + 2²⁰²⁰

2⁴A - A = (2⁴ + 2⁸ + 2¹² + ..... + 2²⁰¹⁶ + 2²⁰²⁰) - (1 + 2⁴ + 2⁸ + ...... + 2²⁰¹² + 2²⁰¹⁶)

15A = 2²⁰¹⁰ - 1

\(\Rightarrow A=\frac{2^{2010}-1}{15}\)

B = 1 + 2² + 2⁴ + ....... + 2²⁰¹⁶ + 2²⁰¹⁸

2²B = 2² + 2⁴ + 2⁶ + ........ + 2²⁰¹⁸ + 2²⁰²⁰

2²B - B = (2² + 2⁴ + 2⁶ + ........ + 2²⁰¹⁸ + 2²⁰²⁰) - (1 + 2² + 2⁴ + ....... + 2²⁰¹⁶ + 2²⁰¹⁸)

3B = 2²⁰¹⁰ - 1

\(\Rightarrow B=\frac{2^{2010}-1}{3}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{2^{2010}-1}{15}}{\frac{2^{2010}-1}{3}}=\frac{\left(2^{2010}-1\right).\frac{1}{15}}{\left(2^{2010}-1\right).\frac{1}{3}}=\frac{\frac{1}{15}}{\frac{1}{3}}=\frac{\frac{1}{3}.\frac{1}{5}}{\frac{1}{3}}=\frac{1}{5}\)

Đáng ra phải là 2²⁰²⁰ chứ bạn

a) ĐKXĐ: \(x\notin\left\{0;2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-1}

bn vt đề hẳn hỏi ra dc ko bn

vt thế này mk ko hiểu nhé

^^

,,,