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\(\hept{\begin{cases}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{cases}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)
<=> x+y = 0 hoặc x+z=0 hoặc z+y=0
<=> x = -y hoặc x = -z hoặc z = -y
\(\Rightarrow P=\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)
\(x-2008=X;y-2009=Y;z-2010=Z\)
\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)
\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)
\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)
\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)
\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)
\(\left\{{}\begin{matrix}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{matrix}\right.\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)
\(\Leftrightarrow x+y=0\) hoặc \(x+z=0\) hoặc \(z+y=0\)
\(\Leftrightarrow x=-y\) hoặc \(x=-z\) hoặc z=-y
\(\Rightarrow P\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)
Chúc bạn học tốt !!
ta nhân cả 2 vế với \(x+\sqrt{x^2+2008}\)
hay \(y+\sqrt{y^2+2008}\)
Lời giải:
Bạn phải thêm đk \(x,y,z\) là những số không âm.
Đặt \(\frac{x}{2008}=\frac{y}{2009}=\frac{z}{2010}=k(k\geq 0)\Rightarrow x=2008k; y=2009k; z=2010k\)
Khi đó:
\(z-x=2010k-2008k=2k\)
\(\left\{\begin{matrix} x-y=2008k-2009k=-k\\ y-z=2009k-2010k=-k\end{matrix}\right.\)
\(\Rightarrow 2\sqrt{(x-y)(y-z)}=2\sqrt{(-k)(-k)}=2\sqrt{k^2}=2|k|=2k\)
Do đó: \(z-x=2\sqrt{(x-y)(y-z)}\)
Ta có đpcm.