\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+.......+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{9}{100}}\)
Chứng minh = 2
Những câu hỏi liên quan
Chứng minh rằng: \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+..+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
Đặt A= 200- (3+\(\frac{2}{3}+\frac{2}{4}+.....+\frac{2}{100}\))
=\(197-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)
=\(\frac{197.2}{2}-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)
=\(2.\left(\frac{196+1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)
=\(2\left(\frac{196}{2}+\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{100}\right)\)
=\(2\left(98+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)
=\(2\left(\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+.....+1-\frac{1}{100}\right)\)
=\(2\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.....+\frac{99}{100}\right)\)
Khi đó \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=\(\frac{2\left(\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=2(đpcm)
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\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{99}{100}}=2\)
chứng minh
Câu này trong đề kì thi Hà NỘi phải ko
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Chứng minh: \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
ta có 200-(3+\(\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\)
=\(1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)
=\(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
thay \(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
ta có \(\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\left(dpcm\right)\)
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Chứng tỏ rằng:
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
Chứng tỏ rằng
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
Chứng tỏ rằng: \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+..........\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+............\frac{99}{100}}\)=2
Chứng tỏ giúp mình với !
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
Ta có \(A=\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{99}{100}}\)
\(A=\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{100}\right)}{\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)}\)
\(A=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\)
\(\Rightarrow A=2\)
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Ủa sao bạn ra được \(\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\) số 2 ở 200 đâu ra vậy ! và \(\frac{3}{2}\)nữa !
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Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{99}{100}\)
Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)
=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)
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\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)
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Chứng minh rằng:a)frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{2010^2}1b)frac{1}{2}+frac{2}{2^2}+frac{3}{2^3}+frac{4}{2^4}+...+frac{100}{2^{100}}2c)frac{1}{3}+frac{2}{3^2}+frac{3}{3^3}+...+frac{100}{3^{100}}frac{3}{4}d)frac{1}{3^3}+frac{1}{4^3}+frac{1}{5^3}+...+frac{1}{n^3}frac{1}{12}left(nin N;nge3right)e)frac{3}{4}+frac{5}{36}+frac{7}{144}+...+frac{2n+1}{n^2left(n+1right)^2}1 (n nguyên dương)g)frac{1}{31}+frac{1}{32}+frac{1}{33}+...+frac{1}{2048}3h)left(frac{2}{1}right)left(frac{4}{3}rig...
Đọc tiếp
Chứng minh rằng:
a)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\)<1
b)\(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)<2
c)\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
d)\(\frac{1}{3^3}+\frac{1}{4^3}+\frac{1}{5^3}+...+\frac{1}{n^3}\)<\(\frac{1}{12}\)\(\left(n\in N;n\ge3\right)\)
e)\(\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)<1 (n nguyên dương)
g)\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{2048}\)>3
h)\(\left(\frac{2}{1}\right)\left(\frac{4}{3}\right)\left(\frac{6}{5}\right)...\left(\frac{200}{199}\right)\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
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