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Ta có : \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Leftrightarrow...+2\left(\dfrac{xyc}{abc}+\dfrac{xbz}{abc}+\dfrac{ayz}{abc}\right)=1\) ( 1 )
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\) ( 2 )
Thế ( 2 ) vào ( 1 ) , ta có :
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=1\)
Ta có: \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)\)
Mặt khác, \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) => \(\frac{ayz+bxz+cxy}{xyz}=0\)=> ayz + bxz + cxy = 0
=> \(\frac{ayz+bxz+cxy}{abc}=0\) => \(\frac{yz}{bc}+\frac{xz}{ac}+\frac{xy}{ab}=0\)
Do đó, \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)=0\)
=> đpcm
Từ a/x + b/y + c/z = 1 =>(a/x + b/y + c/z)^2 = 1
hay a^2/x^2+b^2/y^2+c^2/z^2 +2ab/xy+2ac/xz+2bc/yz =1
(cái này là hằng đẳng thức chắc em biết rồi)
<=>a^2/x^2+b^2/y^2+c^2/z^2=1- 2.(ab/xy+ac/xz+bc/yz)....(1)
Ta lại có
(ab/xy+ac/xz+bc/yz) =(abc/xyz .z/c+abc/xyz .y/b+abc/xyz.x/a)
=abc/xyz .(z/c+y/b+x/a)=0...............(2)
a, \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(a+c\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\) (đpcm)
b, Từ \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\) hay ayz+bxz+cxy=0
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{cxy+ayz+bzx}{abc}=1\)
Mà ayz+bxz+cxy=1
=>\(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\) (đpcm)
sửa lại Mà ayz+bzx+cxy=0 nhé