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Ta có:
\(\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{10}{10.110}\right)x=\)\(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.202}+\frac{100}{3.303}+...+\frac{100}{10.101}\right)x=\)\(10.\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=\)\(10.\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{110}\right)\right]x\)\(=\)\(10.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow\left[ \left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]x=\)\(10.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow x=10\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/149 - 1/150
A = (1 + 1/3 + 1/5 + ... + 1/149) - (1/2 + 1/4 + 1/6 + ... + 1/150)
A = (1 + 1/2 + 1/3 +1/4 + 1/5 + 1/6 + ... + 1/149 + 1/150 - 2.(1/2 + 1/4 + 1/6 + ... + 1/150)
A = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/149 + 1/150) - (1 + 1/2 + 1/3 + ... + 1/75)
A =1/76 + 1/77 + 1/78 + ... + 1/150
=> A/B = 1
Ta có :
\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-1-1-1\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
Thay \(a+b+c=2001\)và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10};\)có :
\(A=2001.\frac{1}{10}-3\)
\(=200,1-3\)
\(=197,1\)
Vậy \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=197,1\)
A = 1/2 + 1/12 + 1/30 + ... + 1/2450
A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50
A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/49 - 1/50
A = (1 + 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + 1/6 + ... + 1/50)
A = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - 2.(1/2 + 1/4 + 1/6 + ... + 1/50)
A = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - (1 + 1/2 + 1/3 + ... + 1/25)
A = 1/26 + 1/27 + 1/28 + ... + 1/50 = B
=> A:B = 1
\(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+....+\frac{1}{2450}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}=B\)
Vậy A = B
Từ a+b+c=2010
\(\Rightarrow\)a= 2010-(b+c)
\(\Rightarrow\)b= 2010-(c+a)
\(\Rightarrow\)c= 2010-(a+b)
Thay vào A, ta được:
A=\(\frac{2010-\left(b+c\right)}{b+c}\)+ \(\frac{2010-\left(c+a\right)}{c+a}\) + \(\frac{2010-\left(a+b\right)}{a+b}\)
A= \(\frac{2010}{b+c}\)+ \(\frac{2010}{c+a}\)+\(\frac{2010}{a+b}\)- 3
A= 2010( \(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)+\(\frac{1}{a+b}\) ) -3
A= 2010. \(\frac{1}{10}\)-3
A=201-3
A= 198
Vậy A=198