Tính:
\(\left(\frac{1000}{1}+\frac{999}{2}+\frac{998}{3}+\frac{997}{4}+...+\frac{2}{999}+\frac{1}{1000}\right)\)\(:\)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1000}\right)\)
Những câu hỏi liên quan
Tính giá trị biểu thức bằng cách thuận tiện nhất:a) frac{3^4-1^2}{4^3-2^1}+frac{7^8-5^6}{8^7-6^5}+...+frac{995^{996}-993^{994}}{996^{995}-994^{993}}+frac{999^{1000}-997^{998}}{1000^{999}-998^{997}}b)frac{4^3}{3^4}-frac{2^1}{1^2}+frac{8^7}{7^8}-frac{6^5}{5^6}+...+frac{996^{995}}{995^{996}}-frac{994^{993}}{993^{994}}+frac{1000^{999}}{999^{1000}}-frac{998^{997}}{997^{998}}c)frac{3^4}{4^3}-frac{1^2}{2^1}+frac{7^8}{8^7}-frac{5^6}{6^5}+...+frac{995^{996}}{996^{995}}-frac{993^{994}}{994^{993}}+frac{999...
Đọc tiếp
Tính giá trị biểu thức bằng cách thuận tiện nhất:
a) \(\frac{3^4-1^2}{4^3-2^1}+\frac{7^8-5^6}{8^7-6^5}+...+\frac{995^{996}-993^{994}}{996^{995}-994^{993}}+\frac{999^{1000}-997^{998}}{1000^{999}-998^{997}}\)
b)\(\frac{4^3}{3^4}-\frac{2^1}{1^2}+\frac{8^7}{7^8}-\frac{6^5}{5^6}+...+\frac{996^{995}}{995^{996}}-\frac{994^{993}}{993^{994}}+\frac{1000^{999}}{999^{1000}}-\frac{998^{997}}{997^{998}}\)
c)\(\frac{3^4}{4^3}-\frac{1^2}{2^1}+\frac{7^8}{8^7}-\frac{5^6}{6^5}+...+\frac{995^{996}}{996^{995}}-\frac{993^{994}}{994^{993}}+\frac{999^{1000}}{1000^{999}}-\frac{997^{998}}{998^{997}}\)
Không sao đâu,các bạn có thể giải từng câu một nhưng phải nhanh lên nhé!
(Các bạn nhớ ghi cách làm nhé!)
:)) ko bt làm :))
kí tên
cái nịt
Tính nhanh :
\(D=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{999}\right).\left(1-\frac{1}{1000}\right)\)
\(D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{998}{999}.\frac{999}{1000}\)
\(D=\frac{1}{1000}\)( rút gọn những thừa số giống nhau ở tử và mẫu)
Vậy \(D=\frac{1}{1000}\)
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D = \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{999}{1000}\)
D = \(\frac{1}{1000}\)
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\(D=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{999}\right)\left(1-\frac{1}{1000}\right)\)
\(\Rightarrow D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{998}{999}.\frac{999}{1000}\)
\(\Rightarrow D=\frac{1.2.3.....998.999}{2.3.4.....999.1000}\)
\(\Rightarrow D=\frac{1}{1000}\)
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chứng minh rằng \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
Áp dụng tính \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
\(VT=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2-\left(\frac{2}{ab}-\frac{2}{a\left(a+b\right)}-\frac{2}{b\left(a+b\right)}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2-\frac{2\left(a+b\right)-2b-2a}{ab\left(a+b\right)}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|=VP\)
Áp dụng tính M: \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
\(M=999.\sqrt{\frac{1}{999^2}+\frac{1}{1^2}+\frac{1}{\left(999+1\right)^2}}+\frac{999}{1000}\)
\(M=999.\left(\frac{1}{1}+\frac{1}{999}-\frac{1}{1000}\right)+\frac{999}{1000}\)
\(M=999+1-\frac{999}{1000}+\frac{999}{1000}=1000\)
Vậy M=1000.
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Tính nhanh:\(\frac{\left(1+2\right)\times3}{\left(2+3\right)\times4}+\frac{\left(2+3\right)\times4}{\left(3+4\right)\times5}+...+\frac{\left(999+1000\right)\times1001}{\left(1000+1001\right)\times1002}+\frac{\left(1000+1001\right)\times1002}{\left(1001+1002\right)\times1003}\)
$\frac{999}{1000}+\frac{998}{1000}+\frac{997}{1000}+...+\frac{1}{1000}$
CMR \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
Áp dụng tính : \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
- Gỉa sử \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}=\left(\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\right)^2\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}+\frac{2}{ab}-\frac{2}{b\left(a+b\right)}-\frac{2}{a\left(a+b\right)}\)
=> \(\frac{2}{ab}-\frac{2}{b\left(a+b\right)}-\frac{2}{a\left(a+b\right)}=0\)
=> \(\frac{a+b}{ab\left(a+b\right)}-\frac{a}{ab\left(a+b\right)}-\frac{b}{ab\left(a+b\right)}=0\)
=> \(\frac{a+b-a-b}{ab\left(a+b\right)}=\frac{0}{ab\left(a+b\right)}=0\) (Luôn đúng )
Vậy ....
- Áp dụng : \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
=> \(M=\sqrt{1+999^2+\frac{999^2}{\left(1+999\right)^2}}+\frac{999}{1000}\) ( với \(a=1,b=999\) )
=> \(M=1+999-\frac{999}{1000}+\frac{999}{1000}=1000\)
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Tính\(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+............+\frac{1}{998}.\frac{1}{999}+\frac{1}{999}.\frac{1}{1000}\)
=1/1*2+1/2*3+...+1/999*1000
=1/1-1/2+1/2-1/3+...+1/999-1/1000
=1-1/1000
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So sánh A và B biết;
A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{999}{1000}\)
B = \(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{998}{999}\)
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\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}+\frac{1}{1000}\)
\(=1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{999}+\frac{1}{999}\right)-\frac{1}{1000}\)
\(=1+0+0+...+0-\frac{1}{1000}\)
\(=1-\frac{1}{1000}=\frac{999}{1000}\)
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Xem thêm câu trả lời
Tính A biết \(A=\frac{1000}{1}+\frac{999}{2}+\frac{998}{3}+...+\frac{2}{999}+\frac{1}{1000}\)
chứng minh rằng
\(\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}}=\left|\frac{1}{x}+\frac{1}{y}-\frac{1}{x+y}\right|\).áp dụng tính M=\(\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
https://i.imgur.com/V1B5vL0.jpg
https://i.imgur.com/cgd6dWC.jpg