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\(\left(ad+bc\right)\left(a^2d^2+b^2c^2\right)=0\)
\(\Rightarrow a^3d^3+adb^2c^2+bca^2d^2+b^3c^3=0\)
\(\Rightarrow a^3d^3+abcd\left(bc+ad\right)+b^3c^3=0\)
\(\Rightarrow a^3d^3+abcd.0+b^3c^3=0\)
\(\Rightarrow a^3d^3+b^3c^3=0\)
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\(a,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0.abc=0\)
Mà \(a+b+c=1=>\left(a+b+c\right)^2=1=>a^2+b^2+c^2+2ab+2bc+2ac=1\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=1=>a^2+b^2+c^2=1-0=1\) (vì ab+bc+ac=0)
\(b,S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3\)
\(=2014.\frac{1}{2014}-3=1-3=-2\)
Vậy.....................
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Ta có :
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017.\frac{1}{10}\)
\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)
Mà \(2017=a+b+c\)nên :
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{b}{a+c}\right)=201,7\)
\(3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7\)
\(\Leftrightarrow M=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7-3\)
\(\Leftrightarrow M=198,7\)
Vậy ...
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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)=>\(\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)
=>abc=(bc+ac+ab)(a+b+c)=ab2+a2b+ac2+a2c+bc2+bc2+3abc
=ab(a+b)+ac(a+c)+bc(b+c)+3abc
=>ab(a+b)+ac(a+c)+bc(b+c)+2abc=0
=>ab(a+b+c-c)+ac(a+c+c-c)+bc(b+c)+2abc=0
=>(a-c)[ac+ab)]+(b+c)(ab+bc)+2ac2+2abc=0
=>(a-c)a(c+b)+(b+c)b(a+c)+2ac(b+c)=0
=>(b+c)[(a-c)a+b(a+c)+2ac]=0
=>(b+c)(a2-ac+ab+bc+2ac)=0
=>(b+c)(a2+ab+bc+ac)=0
=>(b+c)[a(a+b)+c(a+b)]=0
=>(b+c)(a+c)(a+b)=0
*A=(b+c)(a+c)(a+b)+9=0+9=9.