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\(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \Rightarrow P+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\\ \Rightarrow P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\\ =\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=2018.\dfrac{2021}{4034}=1011.000992\\ \Rightarrow P=1008.000992\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
\(a+b+c=2016\Rightarrow\left\{{}\begin{matrix}a=2016-\left(b+c\right)\\b=2016-\left(c+a\right)\\c=2016-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{2016-\left(b+c\right)}{b+c}+\dfrac{2016-\left(c+a\right)}{c+a}+\dfrac{2016-\left(a+b\right)}{a+b}\)\(\Rightarrow S=2016\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(\Rightarrow S=2016.\dfrac{1}{90}-3\)
\(\Rightarrow S=\dfrac{97}{2}\)
Ta có: a+b-c/c = b+c-a/a = c+a-b/b = a+b-c+b+c-a+c+a-b/c+a+b
= a+b+c/a+b+c = 1 (Áp dụng tính chất dãy tỉ số bằng nhau)
Trường hợp 1 : Nếu a+b+c = 0 => a=0; b=0 ; c=0 => P =1
Trường hợp 2: Nếu a+b+c khác 0 => a+b+c = 1
=> a+b = 1-c => b+c = 1-a
=> a+c = 1-b
Ta lại có:
1-c-c/c =1 => 1- 2c/c =1 => 1-2c = c => 1 = 3c=> c= 1/3
1-a-c/a = 1 => 1- 2a/a=1 => 1-2a =a => 1 = 3a => a= 1/3
1-b-b/b = 1 => 1-2b/b = 1 => 1-2b = b => 1= 3b => b= 1/3
=> P= (1+ 1/3 : 1/3). (1+ 1/3 : 1/3). ( 1+ 1/3 :1/3)
= 2 . 2. 2 =8
Vậy P = 1 hoặc = 8
Ta có :
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}:\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}\cdot\dfrac{2}{1}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{b}{ab}+\dfrac{a}{ab}=\dfrac{2}{c}\)
\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{2}{c}\)
\(\Rightarrow2ab=\left(a+b\right)c\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=ac+bc\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a.\left(c-b\right)=b.\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)( đpcm )
\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{c+b}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+c+b}{c+b}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3\)
\(=4034.\dfrac{1}{2}-3=2014\)
Guể?
\(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=\dfrac{4034}{2}=2017\)
\(\Rightarrow1+\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}=2017\)
\(\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=2014\)