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chú ý:1/2=1/2^1
BIỂU THỨC A có:2018-1+1=2018 số hạng
A=(1/2^1+1/2^2018)+(1/2^2+1/2^2017)+...+(1/2^1008+1/2^1011)+(1/2^1009+1/2^1010)
A= 1 + 1 + ... + 1 + 1 (có 2018:2=1009 số 1)
A= 1009
MÌNH GIẢI ĐÊN ĐÂY PHÀN YÊU CẦU CHỨNG MINH BẠN GHI RÕ HỘ MÌNH RỒI MÌNH SẼ GIÚP BẠN TIẾP
đó là việc của bạn quang chứ có liên quan đến bn đâu mà bn hỏi
a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\) => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) => \(\frac{a}{b}=\frac{13}{6}\)
b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)
a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)
<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)
<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)
<=>\(\frac{a}{b}=\frac{13}{6}\)
b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3
=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3
=\(\frac{23}{4}\).\(\frac{5}{3}\)-3
=\(\frac{115}{12}\)-3
=\(\frac{115-36}{12}\)
=\(\frac{79}{12}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2018}}\)
\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2017}}\)
\(2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{2017}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2018}}\right)\)
\(A=1-\dfrac{1}{2^{2018}}\)
Biểu thức cần c/m là : \(2^{2018}\cdot\left(1-\dfrac{1}{2^{2018}}\right)+1\)
\(=2^{2018}-\dfrac{2^{2018}}{2^{2018}}+1\)
\(=2^{2018}-1+1\)
\(=2^{2018}\)
p/s: ko biết biểu thức cần c/m làm sao nữa ? đề hơi thiếu
1) \(\frac{3^{2014}.8^{19}}{6^{60}.3^{1955}}=\frac{3^{2014}.\left(2^3\right)^{19}}{\left(2.3\right)^{60}.3^{1955}}=\frac{3^{2014}.2^{57}}{2^{60}.3^{2015}}=\frac{1}{2^3.3}=\frac{1}{24}\)
2) \(5^x+5^{x+1}=150\)
=> 5x(1 + 5) = 150
=> 5x.6 = 150
=> 5x = 25
=> \(x=\pm2\)
3) \(\frac{3}{11.16}+\frac{3}{16.21}+...+\frac{3}{61.66}=\frac{3}{5}\left(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\right)\)
\(=\frac{3}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\right)=\frac{3}{5}.\left(\frac{1}{11}-\frac{1}{66}\right)=\frac{3}{5}.\frac{5}{66}=\frac{1}{22}\)
1) Đặt \(A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{99}.3\)
Vì \(3⋮3\) nên \(2.3+2^3.3+...+2^{99}.3⋮3\)
hay \(A⋮3\)(đpcm)
2) Đặt \(B=3+3^2+3^3+...+3^{1998}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{1996}+3^{1997}+3^{1998}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{1996}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{1996}.13\)
\(=39+3^3.39+...+3^{1995}.39\)
Vì \(39⋮39\)nên \(39+3^3.39+...+3^{1995}.39⋮39\)
hay \(B⋮39\)(đpcm)
a) 2+22+23+...+2100
=(2+22+23+24+25)+(26+27+28+29+210)+.....+(296+297+298+299+2100)
=2(1+2+22+23+24)+26(1+2+22+23+24)+....+296(1+2+22+23+24)
=2(1+2+4+8+16)+26(1+2+4+8+16)+....+296(1+2+4+8+16)
=2.31+26.31+....+296.31
=31(2+26+....+296)
=> đpcm