Rút gọn tính giá trị : 8/(1.2.3)+8/(2.3.4)+..........+4/49.99.100
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\(a,M=x^2-64+x^3+64=x^3+x^2\\ b,x=-4\Leftrightarrow M=-64+16=-48\\ c,M=0\Leftrightarrow x^2\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)
\(\Rightarrow B=\frac{189}{760}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
tham khảo:
\(P=\)\(\dfrac{8}{x^2+4x}\)+\(\dfrac{5}{x+4}\) - \(\dfrac{2}{x}\)
\(=\) \(\dfrac{8}{x\left(x+4\right)}\)+\(\dfrac{5x}{x\left(x+4\right)}\) - \(\dfrac{2\left(x+4\right)}{x\left(x+4\right)}\)
\(=\) \(\dfrac{8+5x-2x-8}{x\left(x+4\right)}\)
\(=\) \(\dfrac{3}{x\left(x+4\right)}\)
\(=\) \(\dfrac{3}{x+4}\)
Khi \(x =\) \(\dfrac{1}{2}\) thì \(P=\) \(\dfrac{3}{\dfrac{1}{2}+4}\) \(=\) \(\dfrac{2}{3}\)
A=1.2.3+2.3.4+...+2416.2417.2418=(2416.2417.2418.2419):4
B=1.2.3-2.3.4+...+2415.2416.2417-2416.2417.2418
A+B=2(1.2.3+3.4.5+...+2415.2416.2417)=2C
Xét C=1.2.3+3.4.5+...+2415.2416.2417
=1.3(5-3)+3.5(7-3)+...+2415.2417.(2419-3)
=1.3.5+3.5.7+...+2415.2417.2419-3(1.3+3.5+...2415.2417)
=(1.3.5.7+3.5.7.(9-1)+...+2415.2417.2419.(2421-2413)):8-3.(1.3+1.3.5+3.5.(7-1)+...+2415.2417(2419-2413)):6
=2415.2417.2419.2421:8-3.(1.3+2415.2417.2419):6
=> B=2C-A
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Đặt biểu thức trên = A
Xét : B = 1.2.3+2.3.4+....+n.(n+1).(n+2)
4B = 1.2.3.4+2.3.4.4+....+n.(n+1).(n+2).4
= 1.2.3.4+2.3.4.(5-1)+....+n.(n+1).(n+2).[(n+3)-(n-1)]
= 1.2.3.4+2.3.4.5-1.2.3.4+....+n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)
= n.(n+1).(n+2).(n+3)
=> B = n.(n+1).(n+2).(n+3)/4
=> A = 222315.222316.222317.222318/4
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