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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)
\(=\frac{a+ab+1}{a+ab+1}=1\)
Ta có: \(Q+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{b+c}\right)+\left(1+\frac{c}{a+b}\right)\)
\(Q+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(Q+3=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(Q+3=2028\cdot\frac{1}{3}=676\)
=> Q = 676 - 3 = 673
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)
Xét a+b+c=0
\(\Rightarrow a+b=-c,c+b=-a,a+c=-b\)
\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
Xét a+b+c\(\ne0\)
\(\Rightarrow a+b=2c,b+c=2a,c+a=2b\)
\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{2c}{b}\cdot\frac{2a}{c}\cdot\frac{2b}{a}=8\)
Giải:
+) Xét a + b + c = 0
\(\Rightarrow-a=b+c\)
\(\Rightarrow-b=a+c\)
\(\Rightarrow-c=a+b\)
Ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{-c}{c}=\frac{-a}{a}=\frac{-b}{b}=-1\)
Lại có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=-1\)
+) Xét \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Ta có:
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)
Vậy M = -1 hoặc M = 8
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
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
Ta có S + 4 = \(\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{c+d+a}+1\right)+\left(\frac{c}{a+b+d}+1\right)+\left(\frac{d}{a+b+c}+1\right)\)
\(=\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{b+c+d}\)
\(=\left(a+b+c+d\right)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}\right)\)
\(=4000.\frac{1}{40}=100\)(a + b + c + d = 4000 ; \(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}=\frac{1}{40}\))
=> S = 100 - 4 = 96