\(\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2}}}}\)
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23 tháng 8 2023 lúc 20:35
Sdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{9^2}dfrac{1}{2.3}+dfrac{1}{3.4}+dfrac{1}{4.5}...+dfrac{1}{9.10} dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{9^2}dfrac{1}{2}-dfrac{1}{3}+dfrac{1}{3}-dfrac{1}{4}+dfrac{1}{4}-dfrac{1}{5}+...+dfrac{1}{9}-dfrac{1}{10} dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{9^2}dfrac{1}{2}-dfrac{1}{10} dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{9^2}dfrac{2}{5}Sdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{9^2} df...
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S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}...+\dfrac{1}{9.10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2}-\dfrac{1}{10}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{2}{5}\)
S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9}< 1-\dfrac{1}{9}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{8}{9}\)
Vậy \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
Đặt $ X a - b; Y b - c; Z c - a Rightarrow X + Y + Z 0$Với X + Y + Z 0, ta chứng minh được :$ ( dfrac{1}{X} + dfrac{1}{Y} + dfrac{1}{Z} )^2 dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2}$Thật vậy, ta có :$ ( dfrac{1}{X} + dfrac{1}{Y} + dfrac{1}{Z} )^2 dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2} + dfrac{2}{XY} + dfrac{2}{YZ} + dfrac{2}{ZX}$$ dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2} + 2.dfrac{X + Y + Z}{XYZ}$$ dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2}$ ( do X + Y + Z 0)$ Righta...
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Đặt $ X = a - b; Y = b - c; Z = c - a \Rightarrow X + Y + Z = 0$
Với X + Y + Z = 0, ta chứng minh được :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$
Thật vậy, ta có :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + \dfrac{2}{XY} + \dfrac{2}{YZ} + \dfrac{2}{ZX}$
$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + 2.\dfrac{X + Y + Z}{XYZ}$
$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$ ( do X + Y + Z = 0)
$ \Rightarrow \sqrt{\dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}} = \sqrt{( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2} = |\dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z}|$
Suy ra : $ \sqrt{\dfrac{1}{(a - b)^2} + \dfrac{1}{(b - c)^2} +\dfrac{1}{( c - a)^2}} = |\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$
Do a, b, c là số hữu tỷ nên $|\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$ cũng là số hữu tỷ. Ta có điều phải chứng minh.
23 tháng 4 2021 lúc 13:31
a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)
tìm x,y viết dưới dạng phân số
a. 5+dfrac{x}{5+dfrac{2}{5+dfrac{3}{5+dfrac{4}{5}}}}dfrac{x}{1+dfrac{5}{2+dfrac{4}{3+dfrac{3}{5+dfrac{1}{6}}}}}
b.
dfrac{y}{3+dfrac{5}{2+dfrac{4}{2+dfrac{5}{2+dfrac{4}{2+dfrac{5}{3}}}}}}+dfrac{y}{7+dfrac{1}{3+dfrac{1}{3+dfrac{1}{4}}}}
2
c,
x.left(dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+1}}}}}}}}right)2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2}}}}}}}+x.left(3+dfrac{1}{3-df...
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tìm x,y viết dưới dạng phân số
a. \(5+\dfrac{x}{5+\dfrac{2}{5+\dfrac{3}{5+\dfrac{4}{5}}}}=\dfrac{x}{1+\dfrac{5}{2+\dfrac{4}{3+\dfrac{3}{5+\dfrac{1}{6}}}}}\)
b.
\(\dfrac{y}{3+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{3}}}}}}+\dfrac{y}{7+\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{4}}}}\)
= 2
c,
\(x.\left(\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+1}}}}}}}}\right)=\)\(2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}}}}\)+\(x.\left(3+\dfrac{1}{3-\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{3-\dfrac{1}{3}}}}}\right)\)
Giair bằng máy tính casio
bài này đúng là thị của phi...vô của lí ... :))
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25 tháng 4 2023 lúc 9:53
Chứng tỏ rằng:
a)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{3}{4}\)
b)\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
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Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2
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CMR: \(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)là 1 số hữu tỉ
Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)
\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
=>A là số hữu tỉ (ĐPCM)
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2. Chứng minh
a, dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{50^2} 1
b, dfrac{1}{3} dfrac{1}{101}+dfrac{1}{102}+dfrac{1}{103}+...+dfrac{1}{150} dfrac{1}{2}
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2. Chứng minh
a, \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{50^2}\) < 1
b, \(\dfrac{1}{3}\)< \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{150}\)< \(\dfrac{1}{2}\)
Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
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2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
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2b.
Gọi tổng trên là $T$
Chứng minh vế đầu tiên:
Ta có:
$\frac{1}{101}> \frac{1}{150}$
$\frac{1}{102}> \frac{1}{150}$
....
$\frac{1}{149}> \frac{1}{150}$
$\Rightarrow T> \underbrace{\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}}_{50}=\frac{50}{150}=\frac{1}{3}$ (đpcm)
Chứng minh vế số 2:
$\frac{1}{101}< \frac{1}{100}$
$\frac{1}{102}< \frac{1}{100}$
....
$\frac{1}{150}< \frac{1}{100}$
$\Rightarrow T< \underbrace{\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}}_{50}=\frac{50}{100}=\frac{1}{2}$ (đpcm)
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CMR : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Với \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)
\(\dfrac{1}{1\cdot2}>\dfrac{1}{2^2}>\dfrac{1}{2\cdot3},\dfrac{1}{2\cdot3}>\dfrac{1}{3^2}>\dfrac{1}{3\cdot4},...,\dfrac{1}{8\cdot9}>\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(\Rightarrow1-\dfrac{1}{9}>A>\dfrac{1}{2}-\dfrac{1}{10}\) \(\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\)
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CMR:
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)là 1 số hữu tỉ
bạn chứng minh bài toán tổng quát : \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=1+\frac{1}{a}-\frac{1}{a+1}\)rồi áp dụng vào giải bài này nhé
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B(1-dfrac{1}{2^2})(1-dfrac{1}{3^2})(1-dfrac{1}{4^2})(1-dfrac{1}{5^2})...(1-dfrac{1}{99^2})(1-dfrac{1}{100^2})C(dfrac{1}{4}-1)(dfrac{1}{9}-1)(dfrac{1}{16}-1)...(dfrac{1}{999}-1)
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B=(1-\(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))(1-\(\dfrac{1}{5^2}\))...(1-\(\dfrac{1}{99^2}\))(1-\(\dfrac{1}{100^2}\))
C=(\(\dfrac{1}{4}\)-1)(\(\dfrac{1}{9}\)-1)(\(\dfrac{1}{16}\)-1)...(\(\dfrac{1}{999}\)-1)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
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