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cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
hãy tính \(B=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}\)
\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
Mà \(xy+yz+xz=0\)
\(\Rightarrow x^2+y^2+z^2+2.0=0\)
\(\Rightarrow x^2+y^2+z^2=0\)
Mà \(x^2\ge0\)
\(y^2\ge0\)
\(z^2\ge0\)
\(\Rightarrow x^2+y^2+z^2\ge0\)
Mà \(x^2+y^2+z^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
\(\Rightarrow B=\left(0-1\right)^{2007}+0^{2008}+\left(0+1\right)^{2009}\)
\(=\left(-1\right)^{2007}+0+1^{2009}\)
\(=-1+0+1\)
\(=0\)
Vậy ...
\(x+y+z=0< =>\left(x+y+z\right)^2=0< =>x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)
\(< =>x^2+y^2+z^2=0< =>x=y=z=0\)
\(B=\left(-1\right)^{2007}+0+1^{2009}=0\)
x+y+z=0
\(\Rightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow x^2+y^2+z^2=0\)( vì xy+yz+zx=0)
Mà \(x^2+y^2+z^2\ge0\forall x,y,z\Rightarrow x=y=z=0\)
\(\Rightarrow B=\left(0-1\right)^{2007}+0^{2008}+\left(0+1\right)^{2009}\)
= -1+0+1=0
Vậy B=0
Vì x+y+z=0;xy+yz+xz=0
⇒(x+y+z)2=x2+y2+z2+2(xy+yz+xz)=0
⇒(x+y+z)2=x2+y2+z2=0
⇒x=y=z=0
⇒S=(x−1)2005+(y−1)2006+(z+1)2007=(−1)2005+(−1)2006+12007=1
Ta có :\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=x^2+y^2+z^2=0\) (do xy + yz + xz = 0)
Ta lại thấy \(x^2;y^2;z^2\ge0\forall x;y;z\) nên \(x^2+y^2+z^2\ge0\forall x;y;z\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=0\) thay vào S ta được :
\(S=\left(-1\right)^{2005}+\left(-1\right)^{2006}+1^{2007}=1\)
\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
Lời giải:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=0^2-2.0=0$
Ta thấy: $x^2\geq 0; y^2\geq 0; z^2\geq 0$ với mọi $x,y,z\in\mathbb{R}$
Do đó để tổng $x^2+y^2+z^2=0$ thì: $x^2=y^2=z^2=0$
$\Rightarrow x=y=z=0$.
Khi đó:
$B=(0-1)^{2007}+0^{2008}+(0+1)^{2009}$
$=(-1)^{2007}+0+1^{2009}=-1+0+1=0$