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Ta có:
\(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)
\(A=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
\(A=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
Ta lại có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{ab}.\dfrac{1}{-c}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\left(1\right)\)
Thay (1) vào A ta được:
\(A=abc.\dfrac{3}{abc}\)
\(A=3\)
\(s=\frac{bc}{bc\left(1+a+ab\right)}+\frac{1}{1+b+bc}+\frac{b}{b\left(1+c+ac\right)}=>\) \(s=\frac{bc}{bc+abc+ab^2c}+\frac{1}{1+b+bc}+\frac{b}{b+bc+abc}\)=>
\(s=\frac{bc}{1+b+bc}+\frac{1}{1+b+bc}+\frac{b}{1+b+bc}\)=>
\(s=\frac{1+b+bc}{1+b+bc}=1\)Vậy với a.b.c=1 S=1
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 :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
\(M=\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+b}\)
\(\Leftrightarrow M=\dfrac{1}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{a}{a+1+ab}\)
\(\Leftrightarrow M=\dfrac{1+ab+a}{ab+a+1}=1\)
Từ đề bài:A=\(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}=\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}=abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=8\cdot\dfrac{3}{4}=6\)
\(A=\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\)
\(=\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}\\ =abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\\ =8\cdot\dfrac{3}{4}\\ =6\)
Lời giải:
Thay $abc=1$ ta có:
\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)
\(=\frac{c}{c+a.c+ab.c}+\frac{ac}{ac+b.ac+bc.ac}+\frac{1}{1+c+ca}\)
\(=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+ca}=\frac{c+ca+1}{1+c+ca}=1\)