A=(-2).\(\left(-1\dfrac{1}{2}\right)....\left(-1\dfrac{1}{1001}\right)\)
Những câu hỏi liên quan
Tính: \(\left(\dfrac{1000}{1}+\dfrac{999}{2}+\dfrac{998}{3}+...+\dfrac{2}{999}+\dfrac{1}{1000}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1001}\right)\)
Tính: \(\left(\dfrac{1000}{1}+\dfrac{999}{2}+\dfrac{998}{3}+...+\dfrac{2}{999}+\dfrac{1}{1000}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1001}\right)\)
\(\dfrac{1}{2007}.\left(\dfrac{1001}{2006}-2007\right)-\left(\dfrac{1}{2006}-2007\right).\dfrac{1001}{2006}\)=?
1, Tìm các số hữu tỉ:
a) Có dạng dfrac{12}{b} sao cho dfrac{-8}{19} dfrac{12}{b} dfrac{-2}{5}
b) Có dạng dfrac{9}{b} sao cho dfrac{8}{11} dfrac{9}{b} dfrac{12}{13}
2, Tính:
M54-dfrac{1}{2}left(1+2right)-dfrac{1}{3}left(1+2+3right)-dfrac{1}{4}left(1+2+3+4right)-...dfrac{1}{12}left(1+2+3+...+12right)
3, Rút gọn các biểu thức sau:
a) A dfrac{9^9+27^7}{9^6+243^3}
b) B dfrac{left(dfrac{2}{3}right)^5.left(dfrac{-27}{8}right)^2.729}{left(dfrac{3}{2}right)^4.216}
4, Cho a,b,c là các số nguyên dư...
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1, Tìm các số hữu tỉ:
a) Có dạng \(\dfrac{12}{b}\) sao cho \(\dfrac{-8}{19}< \dfrac{12}{b}< \dfrac{-2}{5}\)
b) Có dạng \(\dfrac{9}{b}\) sao cho \(\dfrac{8}{11}< \dfrac{9}{b}< \dfrac{12}{13}\)
2, Tính:
M=\(54-\dfrac{1}{2}\left(1+2\right)-\dfrac{1}{3}\left(1+2+3\right)-\dfrac{1}{4}\left(1+2+3+4\right)-...\dfrac{1}{12}\left(1+2+3+...+12\right)\)
3, Rút gọn các biểu thức sau:
a) A= \(\dfrac{9^9+27^7}{9^6+243^3}\)
b) B= \(\dfrac{\left(\dfrac{2}{3}\right)^5.\left(\dfrac{-27}{8}\right)^2.729}{\left(\dfrac{3}{2}\right)^4.216}\)
4, Cho a,b,c là các số nguyên dương sao cho mỗi số nhỏ hơn tổng của hai số kia. Chứng minh rằng \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\)
5, Cho A= \(\dfrac{1001}{1000^2+1}+\dfrac{1001}{1000^2+2}+...+\dfrac{1001}{1000^2+1000}\)
Chứng minh rằng 1<A2 < 4
từ giả thiết, ta có dfrac{1}{xy}+dfrac{1}{yz}+dfrac{1}{zx}1
đặt left(dfrac{1}{xy};dfrac{1}{yz};dfrac{1}{zx}right)left(a;b;cright)Rightarrow a+b+c1 left(dfrac{ac}{b};dfrac{ab}{c};dfrac{bc}{a}right)left(dfrac{1}{x^2};dfrac{1}{y^2};dfrac{1}{z^2}right)
ta có VTdfrac{1}{sqrt{1+dfrac{1}{x^2}}}+dfrac{1}{sqrt{1+dfrac{1}{y^2}}}+dfrac{1}{sqrt{1+dfrac{1}{z^1}}}sqrt{dfrac{1}{1+dfrac{ac}{b}}}+sqrt{dfrac{1}{1+dfrac{ab}{c}}}+sqrt{dfrac{1}{1+dfrac{bc}{a}}}
dfrac{1}{sqrt{dfrac{b+ac}{b}}}+dfrac{1}{sqrt{dfrac...
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từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
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Chứng minh rằng :
a)dfrac{1}{x}-dfrac{1}{x+a}dfrac{a}{xleft(x+aright)}
b)dfrac{1}{xleft(x+1right)}-dfrac{1}{left(x+1right)left(x+2right)}dfrac{2}{xleft(x+1right)left(x+2right)}
c)dfrac{1}{xleft(x+1right)left(x+2right)}-dfrac{1}{left(x+1right)left(x+2right)left(x+3right)}dfrac{3}{xleft(x+1right)left(x+2right)left(x+3right)}
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Chứng minh rằng :
a)\(\dfrac{1}{x}\)-\(\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
b)\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\)
c)\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
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a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)
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rút gọn \(\dfrac{1}{2\left(a+b\right)^3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}\right)+\dfrac{3}{2\left(a+b\right)^4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)+\dfrac{3}{\left(a+b\right)^5}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
cái này tương tự này, do dài quá nên ngại làm, bn tham khảo nhé Câu hỏi của Thiên An - Toán lớp 9 - Học toán với OnlineMath
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Cho A=\(\left(\dfrac{1}{2^2}-1\right)\)\(\left(\dfrac{1}{3^2}-1\right)\)\(\left(\dfrac{1}{4^2}-1\right)\)...\(\left(\dfrac{1}{2013^2}-1\right)\)\(\left(\dfrac{1}{2014^2}-1\right)\) và B= \(-\dfrac{1}{2}\)
Hãy so sánh A và B
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
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