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a.
\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)
\(=\left(-\frac{1}{2}\right)\times\left(-\frac{2}{3}\right)\times\left(-\frac{3}{4}\right)\times...\times\left(-\frac{2015}{2016}\right)\times\left(-\frac{2016}{2017}\right)\)
\(=\frac{1}{2017}\)
b.
\(\frac{2^{50}\times7^2+2^{50}\times7}{4^{26}\times112}=\frac{2^{50}\times\left(7^2+7\right)}{\left(2^2\right)^{26}\times112}=\frac{2^{50}\times\left(49+7\right)}{2^{52}\times2\times56}=\frac{56}{2^3\times56}=\frac{1}{8}\)
a. (1/2-1).(1/3-1)(1/4-1). ... .(1/2017-1)=(-1/2)(-2/3)(-3/4). ... .(-2016/2017)
Vì dãy số có 2016 số hạng âm nên tích của chúng là một số dương.
Ta có:(-1/2)(-2/3)(-3/4). ... . (-2016/2017)=1/2017
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) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
a) Vì 2,5 > 2,125 nên -2,5 < -2,125
b) Vì \( - \frac{1}{{10000}}\)< 0 và 0 < \(\frac{1}{{23456}}\)nên \( - \frac{1}{{10000}}\) < \(\frac{1}{{23456}}\)
Chú ý: Số hữu tỉ âm luôn nhỏ hơn số hữu tỉ dương.
\(-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\)
\(=-\frac{10000}{10000}-\frac{1000}{10000}-\frac{100}{10000}-\frac{10}{10000}-\frac{1}{10000}\)
\(=\frac{-10000-1000-100-10-1}{10000}\)
\(=-\frac{11111}{10000}=-1,1111\)
\(=-\left(1+\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}\right)\)
\(=-\left(\frac{10000}{10000}+\frac{1000}{10000}+\frac{100}{10000}+\frac{1}{10000}\right)\)
\(=-\left(\frac{10000+1000+100+10+1}{10000}\right)\)
\(=-\left(\frac{11111}{10000}\right)\)
Vậy.....
\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(x=\frac{1}{100}.\frac{101}{2}\)
\(x=\frac{101}{200}\)
\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)
\(X=\frac{1}{100}.\frac{101}{2}\)
\(X=\frac{101}{200}\)
Study well
\(\frac{3}{2^2}.\frac{2^3}{3^2}.\frac{5.3}{4^2}.......\frac{3.3333}{100^2}\)
mk ko nghĩ ra phần sau chắc là rút gọn thì phải!!?
bạn làm thử xem
Ta có : \(B=\frac{-1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(\Rightarrow B=-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\right)\)
Đặt \(A=\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\)
\(\Rightarrow10A=1+\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}\)
\(\Rightarrow10A-A=1-\frac{1}{100000}\)
\(\Rightarrow9A=\frac{99999}{100000}\)
\(\Rightarrow A=\frac{99999}{100000}.\frac{1}{9}=\frac{11111}{100000}\)
=> B = \(-\frac{11111}{100000}\)
\(A=\frac{-1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(\Rightarrow A=-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000}{100000}+\frac{1000}{100000}+\frac{100}{100000}+\frac{10}{100000}+\frac{1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{10000+1000+100+10+1}{100000}\right)\)
\(\Rightarrow A=-\left(\frac{11111}{100000}\right)\)
\(\Rightarrow A=\frac{-11111}{100000}\)
Ta có:
\(2000x=\frac{2000x}{2000x};5000y=\frac{5000y}{5000y};10000x=\frac{10000z}{10000z}\Rightarrow x=y=z\)
=> 2000x = 5000y = 10000z \(=\frac{2000x}{2000x}=\frac{5000y}{5000y}=\frac{10000z}{10000z}\)
=> \(\frac{2000x}{10000}=\frac{5000y}{10000}=\frac{10000z}{10000}\)
chắc vậy