\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

[(2³ - 5) . (-3)+9] . (2²⁰¹² . 2011 - 2012² . 2011+1)

= [ 3 . ( -3 ) + 9] . (2²⁰¹² . 2011 - 2012² . 2011+1)

=  [ (-9) + 9 ] . (2²⁰¹² . 2011 - 2012² . 2011+1)

= 0 . (2²⁰¹² . 2011 - 2012² . 2011+1)

= 0

[ (2³ - 5) . (-3) + 9 ] . ( 2²⁰¹² . 2011 - 2012² . 2011+1 )

= [ 3 . ( -3 ) + 9] . (2²⁰¹² . 2011 - 2012² . 2011+1 )

= 0 . (2²⁰¹² . 2011 - 2012² . 2011+1) = 0

Đặt N = 1 + 2 + 2² +...+ 2²⁰¹²

2N = 2 + 2² + 2³ +...+ 2²⁰¹³

2N - N = (2 + 2² + 2³+....+ 2²⁰¹³) - (1 + 2 + 2² +....+ 2²⁰¹²)

N = 2²⁰¹³ - 1

Thay N vào M ta được:

Đặt \(N=1+2+2^2+...+2^{2012}\)

\(2N=2+2^2+2^3+...+2^{2013}\)

\(2N-N=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)

\(N=2^{2013}-1\)

Thay N vào M ta được:

\(M=\dfrac{2^{2013-1}}{2^{2014}-2}=\dfrac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\dfrac{1}{2}\)

Tham khảo link: https://olm.vn/hoi-dap/detail/80564627052.html

a: \(61\cdot45+61\cdot23-68\cdot51\)

\(=61\left(45+23\right)-68\cdot51\)

\(=68\cdot61-68\cdot51\)

\(=68\left(61-51\right)=68\cdot10=680\)

b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)

\(=75-75+4\cdot8\)

\(=4\cdot8=32\)

c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)

\(=\dfrac{36}{20-30+4^2}\)

\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)

d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)

\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)

a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)

Vậy: \(1+2^2+2^3+...+2^{10}=2045\)

b) 

a] \(60-3\left(x-1\right)=2^3\cdot3\)

\(\Rightarrow60-3\left(x-1\right)=24\)

\(\Rightarrow3\left(x-1\right)=36\)

\(\Rightarrow x-1=12\)

\(\Rightarrow x=13\)

b] \(\left(3x-2\right)^3=2\cdot2^5\)

\(\Rightarrow\left(3x-2\right)^3=2^6\)

\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)

\(\Rightarrow3x-2=2^2\)

\(\Rightarrow3x=6\)

\(x=2\)

c] \(5^{x+1}-5^x=500\)

\(\Rightarrow5^x\left(5-1\right)=500\)

\(\Rightarrow5^x\cdot4=500\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

d] \(x^2=x^4\)

\(\Rightarrow x=x^2\)

\(\Rightarrow x-x^2=0\)

\(\Rightarrow x\left(1-x\right)=0\)

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

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

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