Đặt biểu thức trên là B

B=-(2^100+2^99+2^98+..+2^2+2+1)

Tự túc nhé

Lời giải:

$\frac{2^{13}+2^{50}}{2^{10}+2}+\frac{2(2^{12}+2^{49})}{2(2^9+1)}$

$=\frac{2^{12}+2^{49}}{2^9+1}$

( 1² + 2² + 3² +....+ 2012² ). (91 - 91) = (1² + 2² + 3² +....+ 2012²) . 0 = 0

 [91-273/3]x[1²+2²+3²+...+2012²]

=[91-91]x[1²+2²+3²+...+2012²]

=0x[1²+2²+3²+...+2012²]

=0

2³.2⁴.2⁶ = 2³⁺⁴⁺⁶= 2¹³

2³ . 2⁴ . 2⁶ = 2³⁺⁴⁺⁶ = 2¹³

\(4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2009\)

\(=4.\left\{9.\left[\left(25^{ }+8\right):11\right]-26\right\}+2009\)

\(=4.\left\{9.\left[33:11\right]-26\right\}+2009\)

\(=4.\left\{9.3-26\right\}+2009\) \(=4.\left(27-26\right)+2009\)

\(=4.1+2009=2013\)

thì khúc này \(\frac{3^{12}.2\left(3+2\right)}{3^{12}.2}\)

Ta thấy Tử và mẫu có \(3^{12}.2\)chung nên ta rút gọn cho nhau thì còn lại \(3+2=5\)

\(\frac{3^{13}.2+3^{12}.2^2}{3^{12}.2}=\frac{3^{12}.2\left(3+2\right)}{3^{12}.2}=3+2=5\)

\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+...+\frac{2}{899}\)
\(=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{29\cdot31}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{31}\)
\(=\frac{1}{3}-\frac{1}{31}\)
\(=\frac{28}{93}\)

\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{899}\)

= \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{29.31}\)

= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)

= \(\frac{1}{3}-\frac{1}{31}+0+0+...+0\)

= \(\frac{29}{93}\)

= (-1/2)^3-2+1.2.3+1

=1/4.6+1

=3/2+1

=5/2