Gỉa sử p và q là các số nguyên sao cho:
\(\dfrac{p}{q}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{1334}+\dfrac{1}{1335}\)
Chứng minh rằng p\(⋮\) 2003
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Gỉa sử p và q là các số nguyên sao cho
\(\dfrac{p}{q}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4} +...-\dfrac{1}{1334}+\dfrac{1}{1335}\)
CMR: p \(⋮\)2003
Giả sử m và n là các số nguyên sao cho:\(\dfrac{m}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{1334}+\dfrac{1}{1335}\) .Chứng minh rằng m chia hết cho 2003
Cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{2003^2}\)
Chứng minh rằng \(\dfrac{1}{3}< A< 1\)
Giả sử m và n là các số nguyên sao cho:\(\frac{m}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1334}+\frac{1}{1335}\).Chứng minh rằng m chia hết cho 2003
dfrac{x+4}{2000}+dfrac{x+3}{2001}dfrac{x+2}{2002}+dfrac{x+1}{2003}
1/* Chứng minh rằng:
dfrac{1}{1times2}+dfrac{1}{3times4}+dfrac{1}{5times6}+...dfrac{1}{49times50}dfrac{1}{26}+dfrac{1}{27}+dfrac{1}{28}+..+dfrac{1}{50}
2/* Cho:
Adfrac{1}{1times2}+dfrac{1}{3times4}+dfrac{1}{5times6}+.....+dfrac{1}{99times100}. Chứng minh rằng:dfrac{7}{12} Adfrac{5}{6}
Các bn giúp mk những bài này nha!
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\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
1/* Chứng minh rằng:
\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+..+\dfrac{1}{50}\)
2/* Cho:
A=\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+.....+\dfrac{1}{99\times100}\). Chứng minh rằng:\(\dfrac{7}{12}< A>\dfrac{5}{6}\)
Các bn giúp mk những bài này nha!
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
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\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
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1/ Ta có :
\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)
\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)
\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)
\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)
Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
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Cho a,b,c là các số nguyên khác 0 thỏa:
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
Chứng minh rằng
\(a^3+b^3+c^3\) chia hết cho 3
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)
Theo đề bài \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Rightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=0\)
\(\Rightarrow\frac{c+a+b}{abc}=0\) mà \(a;b;c\ne0\Rightarrow abc\ne0\Rightarrow a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=-\left(a^3+b^3+c^3\right)\)
Mà \(3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) chia hết cho 3 nên \(-\left(a^3+b^3+c^3\right)\) chia hết cho 3
Nên \(a^3+b^3+c^3\) chia hết cho 3
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a, Cho b là số tự nhiên, b>1. Chứng minh rằng: \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\)
b, Áp dụng phần a: Cho S\(=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng: \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)
\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)
b)Áp dụng từ câu a
=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)
.........................
\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)
=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)
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1. Chứng minh rằng:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100}< 1\)
2. Chứng minh rằng:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}< 2\)
1.
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\)\(\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
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2.
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\)\(\dfrac{1}{100!}\)
Ta có:
\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+...+\)\(\dfrac{99.100}{100!}-\dfrac{1}{100}\)
\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+...+\dfrac{99.100}{100!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)
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Chứng minh rằng:
a) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{2}{63}>2\)
b) \(\dfrac{3}{9\cdot14}+\dfrac{3}{14\cdot19}+\dfrac{3}{19\cdot24}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}< \dfrac{1}{15}\)
c) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
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