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Giả sử x/a=y/b=z/c=k
=>x=ak; y=bk; z=ck
\(\left(ax+by+cz\right)^2\)
\(=\left(a^2k+b^2k+c^2k\right)^2\)
\(=k^2\cdot\left(a^2+b^2+c^2\right)^2\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(=\left(a^2+b^2+c^2\right)\left(k^2a^2+k^2b^2+k^2c^2\right)\)
\(=k^2\left(a^2+b^2+c^2\right)^2\)
Do đó: \(\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
1. \(x^2+y^2+z^2+3=2\left(x+y+z\right)< =>x^2-2x+1+y^2-2y+1+z^2-2z+1=0< =>\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)
=>x-1=0<=>x=1
y-1=0<=>y=1
z-1=0<=>z=1
vậy....
2. \(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)
<=>\(\dfrac{2-x}{2008}+1=\dfrac{1-x}{2009}+1-\dfrac{x}{2010}+1\)
<=>\(\dfrac{2010-x}{2008}=\dfrac{2010-x}{2009}+\dfrac{2010-x}{2010}\)
<=>(2010-x)(1/2008-1/2009-1/2010)=0
vì 1/2008-1/2009-1/2010 khác 0 nên 2010-x=0<=>x=2010
1)\(x^2+y^2+z^2+3=2\left(x+y+z\right)\)
\(\Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)
\(\Leftrightarrow x=y=z=1\)
2)\(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{2-x}{2008}+1=\dfrac{1-x}{2009}+1-\dfrac{x}{2010}+1\)
\(\Leftrightarrow\dfrac{2010-x}{2008}=\dfrac{2010-x}{2009}+\dfrac{2010-x}{2010}\)
\(\Leftrightarrow\left(2010-x\right)\left(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\right)=0\)
\(\Leftrightarrow x=2010\)(vì \(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\ne0\))
Ta có : \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Leftrightarrow2axby+2axvz+2bycz=a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2\)
\(\Leftrightarrow a^2y^2+b^2x^2+a^2z^2+c^2x^2+b^2z^2+c^2y^2-2axby-2azcx-2bycz=0\)
\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Do \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
:D
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
Ta có:(a2+b2+c2)(x2+y2+z2)=(ax+by+cz)2
=>a2x2+a2y2+a2z2+b2x2+b2y2+b2z2+c2x2+
c2y2+c2z2=a2x2+b2y2+c2z2+2axby+2axcz+
2bycz
=>a2y2+a2z2+b2x2+b2z2+c2x2+c2y2-2axby-2axcz-2bycz=0
=>(a2y2-2axby+b2x2)+(a2z2-2axcz+c2x2)+
(b2z2-2bycz+c2y2)=0
=>(ay-bx)2+(az-cx)2+(bz-cy)2=0
Vì (ay-bx)2\(\ge0\);(az-cx)2\(\ge0\);(bz-cy)2\(\ge0\)
nên =>(ay-bx)2+(az-cx)2+(bz-cy)2\(\ge0\)
Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\)=>\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)(x;y;z\(\ne0\))
cái chỗ dấu = xảy ra khi... cậu viết rõ hơn đc k? tớ ms vào nên k biết kí hiệu này lắm
1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)
\(\Rightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=45\)
\(\Rightarrow x^2+y^2-2xy=9\)
\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)
\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)
Vậy...
Giả sử như x/2=y/5=-z/2
Đặt x/2=y/5=z/-2=k
=>x=2k; y=5k; z=-2k
\(\left(2x+5y-2z\right)^2=\left(4k+25k+4k\right)^2=\left(41k\right)^2\)
\(33\left(x^2+y^2+z^2\right)=33\left(4k^2+25k^2-4k^2\right)\)
\(=33\cdot25k^2< >\left(41k\right)^2\)
=>Đề sai rồi bạn