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Lời giải:
Ta có:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Rightarrow x^2+y^2+z^2=\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)(a^2+b^2+c^2)\)
\(\Leftrightarrow x^2+y^2+z^2=x^2+\frac{x^2b^2}{a^2}+\frac{x^2c^2}{a^2}+y^2+\frac{y^2a^2}{b^2}+\frac{y^2c^2}{b^2}+z^2+\frac{z^2a^2}{c^2}+\frac{z^2b^2}{c^2}\)
\(\Leftrightarrow \frac{x^2b^2}{a^2}+\frac{x^2c^2}{a^2}+\frac{y^2a^2}{b^2}+\frac{y^2c^2}{b^2}+\frac{z^2a^2}{c^2}+\frac{z^2b^2}{c^2}=0(*)\)
Bởi vì mỗi số hạng trong tổng $(*)$ đều là những số không âm, cho nên để tổng các số không âm bằng $0$ thì bản thân mỗi số đó phải bằng $0$
Do đó:
\(\Leftrightarrow \frac{x^2b^2}{a^2}=\frac{x^2c^2}{a^2}=\frac{y^2a^2}{b^2}=\frac{y^2c^2}{b^2}=\frac{z^2a^2}{c^2}=\frac{z^2b^2}{c^2}=0\)
Do $a,b,c\neq 0$ nên \(x^2=y^2=z^2=0\Rightarrow x=y=z=0\)
Khi đó:\(T=x^{2016}+y^{2016}+z^{2016}=0\)
Cô ơi em có cách khác ạ :)
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
Dấu "=" xảy ra tại x=y=z=0
Khi đó T=0
Ta có:
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
<=> \(\left(a^2+b^2+c^2\right)\)\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)
<=> \(x^2+y^2+z^2=\left(a^2+b^2+c^2\right)\frac{x^2}{a^2}+\left(a^2+b^2+c^2\right)\frac{y^2}{b^2}+\left(a^2+b^2+c^2\right)\frac{z^2}{c^2}\)
<=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)
vì a, b , c khác 0 nên \(\frac{\left(b^2+c^2\right)}{a^2};\frac{\left(c^2+a^2\right)}{b^2};\frac{\left(b^2+a^2\right)}{c^2}\ne0\)
\(\frac{\left(b^2+c^2\right)}{a^2}x^2\ge0;\frac{\left(a^2+c^2\right)}{b^2}y^2\ge0;\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x, y, z
=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x; y; z
Do đó: \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)
=> x = y = z = 0
Vậy T = 0
vi a/x + b/y + c/z =0 suy ra ayz/xyz + bxz/xyz + cxy/xyz =0 suy ra ayz+bxz+cxy /xyz =0 suy ra ayz + bxz + cxy =0
vi x/a + y/b =z/c =0 suy ra (x/a + y/b + z/c )^2 =0 suy ra x^2/a^2 +y^2/b^2 + z^2/c^2 + 2(xy/ab + xz/ac + yz/bc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(cxy+ bxz +ayz /abc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 +2011 = 2011
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\) =>\(\dfrac{ayz+bxz+cxy}{xyz}=0\) =>\(ayz+bxz+cxy=0\) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\)=>\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ac}\right)=0=>\dfrac{x^2}{a2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+ayz+bxz}{abc}\right)=0\) \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=0\) (vì ayz+bxz+cxy=0)
Vậy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2011=2011\)
a,Sửa lại đề nha bạn:Tính A = \(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcz+acy+abz=0\)
(2) \(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{ac}{xz}+\dfrac{bc}{xz}\right)=4\)\(\Rightarrow A=\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=4-2.\left(\dfrac{abz+acy+bcz}{xyz}\right)=4\)b, \(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow a^2+2ab+b^2=c^2\Rightarrow a^2+b^2-c^2=-2ab\)Tương tự: \(b^2+c^2-a^2=-2bc\)
\(c^2+a^2-b^2=-2bc\)
Vậy \(B=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ac}{-2ac}=\dfrac{-3}{2}\)