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\(A=\)\(\left(1^3-1000\right).\left(2^3-1000\right)\)\(.....\left(2018^3-1000\right)\)
\(A=\left(1^3-1000\right).\left(2^3-1000\right)...\left(10^3-1000\right)...\left(2018^3-1000\right)\)
\(A=\left(1^3-1000\right).\left(2^3-1000\right)...0...\left(2018^3-1000\right)\)
\(A=0\)
~~~Hok tốt~~~
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-10^3\right)..\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-1000\right)..\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...0..\left(1000-15^3\right)}\)
= \(2009^0\)
= \(1\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)
\(\frac{2+3+4+....+33}{2}\)
\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{20}.\left(1+...+20\right).\)
\(=1+\frac{3}{2}+\frac{6}{3}+\frac{10}{4}+...+\frac{210}{20}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=\frac{230}{2}=115\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
Ta thấy trong tích có thừa số ( 216 - 6\(^3\)) = 216 - 216 = 0 nên tích bằng 0
Vậy M = 0 + ( 216 x 64 + 216 x 36 )
M = 216 . ( 64 + 36 )
M = 216 x 100
M = 21600
Trong biểu thức trên có chứa (1000-103), mà (1000-103)=1000-1000=0
Do đó tích trên bằng 0
\(\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)...\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-1000\right)...\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\cdot0\cdot\left(1000-50^3\right)\)
\(=0\)