Tính nhanh :
a ) A = (-1)2n . (-1)n . (-1)n+1
b ) B = (10000-12)(10000-22)(10000-32)...(10000-10002)
c ) C = (1/125-1/13)(1/125-1/23)(1/125-1/33)...(1/125-1/253)
K
Khách
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Những câu hỏi liên quan
NN
21 tháng 8 2020
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
24 tháng 7 2017
\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)
\(A=\left(-1\right)^{2n+n+n+1}\)
\(A=\left(-1\right)^{4n+1}\)
\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)
\(B=0\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=0\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)
\(D=1999^0\)
\(D=1\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
\(log_5125=log_55^3=3\)
\(log_6216=log_66^3=3\)
\(log_{10}\dfrac{1}{10000}=log_{10}10^{-4}=-4\)
\(log\sqrt{1000}=log_{10}10^{\dfrac{3}{2}}=\dfrac{3}{2}\)
\(81^{log_35}=3^{3log_35}=3^{log_3125}=125\)
\(125^{log_52}=5^{3log_52}=5^{log_58}=8\)
\(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}=7^{-2log_7\dfrac{1}{8}}=7^{log_764}=64\)
\(\left(\dfrac{1}{625}\right)^{log_52}=5^{-4log_52}=5^{log_5\dfrac{1}{16}}=\dfrac{1}{16}\)
23 tháng 2 2020
\(M=\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+\frac{37}{1000}+...+\frac{121}{1000}+\frac{133}{1000}\)
\(=\frac{1+13+25+37+...+121+133}{1000}\)
\(=\frac{804}{1000}=\frac{201}{250}\)
NQ
15 tháng 8 2017
a) 5x+x+1=\(\dfrac{125}{25}\)
\(\leftrightarrow\) 52x+1 =51
\(\leftrightarrow\) 2x+1=1
\(\leftrightarrow\)2x=0
\(\leftrightarrow\) x=0
31 tháng 7 2018
a)\(\dfrac{1}{10000}+\dfrac{13}{10000}+\dfrac{25}{10000}+...+\dfrac{97}{10000}+\dfrac{109}{10000}\)
\(=\dfrac{1+13+25+...+97+109}{10000}\)
\(=\dfrac{\left(1+109\right)\left[109-1\right]:12+1}{20000}\)
\(=\dfrac{110.10}{20000}=\dfrac{11}{200}\)
b)\(\dfrac{4}{3}\times2019\times0,75\)
=\(\dfrac{4}{3}\times\dfrac{3}{4}\times2019\)
\(=2019\)
c)\(4\times5\times0,25\times\dfrac{1}{5}\times\dfrac{1}{2}\times2\)
\(=\left(4\times\dfrac{1}{4}\right)\left(5\times\dfrac{1}{5}\right)\left(2\times\dfrac{1}{2}\right)\)
\(=1\times1\times1=1\)
Ý d) đặt tính kiểu gì thế ?
FI
1
PT
24 tháng 8 2018
\(\left(3x-5\right)^8=\frac{1}{125}\left(5-3x\right)^{11}\)
\(\Leftrightarrow-\left(3x-5\right)^8=-\frac{1}{125}\left(3x-5\right)^{11}\)
\(\Leftrightarrow-1=-\frac{1}{125}\left(3x-5\right)^3\)
\(\Leftrightarrow\frac{1}{125}\left(3x+5\right)^3=1\)
\(\Leftrightarrow\left(3x-5\right)^3=125\)
\(\Leftrightarrow3x-5=\sqrt[3]{125}\)
\(\Leftrightarrow3x-5=5\)
\(\Leftrightarrow3x=10\)
\(\Leftrightarrow x=\frac{10}{3}\)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{\frac{10}{3}\right\}\)
Các bạn giúp mn với ^^ mn k cho
Các bạn giúp mn với ^^ mn k cho