Thực hiện phép tính
M=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
K
Khách
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7 tháng 2 2017
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{16}.\frac{16.17}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(=\frac{\frac{17.18}{2}-1}{2}=76\)
JA
2 tháng 3 2016
a/ 2H=2^2011-2^2010-2^2009-...-2
=> 2H-H=2^2011-2^2010-2^2009-...-2-(2^2010-2^2009-2^2008-...-1)
H=2^2011-2^2010-2^2009-...-2-2^2010+2^2009+2^2008+...+1
H=2^2011-2^2010-2^2010-1
H=2^2011-2.2^2010-1
H=2^2011-2^2011-1
H=-1 => 2010^-1=1/2010
b/ M=1 + 1/2(1+2) + 1/3(1+2+3) + 1/4(1+2+3+4) + ... + 1/16(1+2+3+...+16)
M= 1+1/2.(2.3/2) + 1/3.(3.4/2) + 1/4.(4.5/2) + ... + 1/16.(16.17/2)
M= 1 + 3/2 +4/2 + 5/2 + ... + 17/2
Cùng mẫu số rồi Tự tính nhé
có 1 công thức làm bài này nè em : 1+2=3=2.3/2, 1+2+3=6=3.4/2, 1+2+3+4=10=4.5/2 ....
22 tháng 4 2020
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
NN
22 tháng 4 2020
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2019
\(S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(A=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2\)
\(\Rightarrow2A-2^{2010}+1=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2A-2^{2010}+1=A\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow S=2^{2010}-A=2^{2010}-\left(2^{2010}-1\right)=1\)
b/ Ta có công thức \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
Do đó:
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+...+\dfrac{1+2+3+...+16}{16}\)
\(P=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{2.3}+\dfrac{4.5}{2.4}+...+\dfrac{16.17}{2.16}\)
\(P=1+\dfrac{1}{2}\left(3+4+5+...+17\right)\)
\(P=1+\dfrac{1}{2}.\dfrac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)
4 tháng 1 2017
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
8 tháng 4 2022
a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\)
b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\)
c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)
2 tháng 11 2019
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+...+\frac{1}{16}.\frac{16.\left(16+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=\frac{\left(17-2+1\right).\left(17+2\right)}{2}:2\)
\(=76\)
CC
2 tháng 11 2019
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}\left[\frac{\left(2+1\right)2}{2}\right]+\frac{1}{3}\left[\frac{\left(3+1\right)3}{3}\right]+...+\frac{1}{16}\left[\frac{\left(16+1\right)16}{2}\right]\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2+2+1+3+1+...+16+1}{2}\)
\(=\frac{\left(1+1+1+..15cs.+1\right)+\left(2+3+...+16\right)+2}{2}\)
\(=\frac{15+135+2}{2}\)
\(=\frac{152}{2}\)\(=76\)