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B1:
pt <=> \(\dfrac{3x^2}{10}+\dfrac{2y^2}{15}+\dfrac{z^2}{20}=0\)
<=> x = y = z = 0
B2: Áp dụng bđt Cô-si:
\(\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)\ge2+2=4\)
Dấu "=" xảy ra <=> x2 = y2 = 1
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\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{xbc+yac+zab}{abc}=1\\ \Rightarrow xbc+yac+zab=abc\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=\dfrac{ayz+bxz+cxy}{xyz}=0\\ \Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(abc\right)^2}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc+yac+zab\right)^2}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc\left(ayz+bxz+cxy\right)}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc.0}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}=1\)
vậy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)(đpcm)
\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.\left(\dfrac{c}{z}+\dfrac{b}{y}+\dfrac{a}{x}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.0=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
a, \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{x^2+y^2}{4+16}=\dfrac{2000}{20}=100\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=100.4=400\\y^2=100.16=1600\\z^2=100.25=2500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm20\\y=\pm40\\z=\pm50\end{matrix}\right.\)
Do \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}x=20\\y=40\\z=50\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-20\\y=-40\\z=-50\end{matrix}\right.\)
Vậy ...
b, \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}=\dfrac{14-6}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.3=3\\z-3=1.4=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
Vậy ...
c, \(x-z=-2\Rightarrow x+2=z\)
Do đó \(y.z=12\Leftrightarrow y.\left(x+2\right)=12\Rightarrow xy+2y=12\Rightarrow6+2y=12\)
\(\Rightarrow y=3\Rightarrow x.3=6\Rightarrow x=2\Rightarrow2-z=-2\Rightarrow z=4\)
Vậy x=2; y=3; z=4
Lời giải:
\(\frac{x^2+y^2-z^2}{2xy}+\frac{y^2+z^2-x^2}{2yz}+\frac{x^2+z^2-y^2}{2xz}=1\)
\(\Leftrightarrow \frac{x^2+y^2-z^2}{2xy}+1+\frac{y^2+z^2-x^2}{2yz}-1+\frac{x^2+z^2-y^2}{2xz}-1=0\)
\(\Leftrightarrow \frac{(x+y-z)(x+y+z)}{2xy}+\frac{(y-z-x)(y-z+x)}{2yz}+\frac{(x-z-y)(x-z+y)}{2xz}=0\)
\(\Leftrightarrow (x+y-z)\left[\frac{x+y+z}{2xy}+\frac{y-z-x}{2yz}+\frac{x-z-y}{2xz}\right]=0\)
\(\Leftrightarrow (x+y-z)(xz+yz+z^2+xy-zx-x^2+xy-zy-y^2)=0\)
\(\Leftrightarrow (x+y-z)[z^2-(x-y)^2]=0\Leftrightarrow (x+y-z)(z-x+y)(x+z-y)=0\)
Nếu $x+y-z=0$ thì:
\(\frac{x^2+y^2-z^2}{2xy}=\frac{(x+y)^2-z^2-2xy}{2xy}=-1\); \(\frac{y^2+z^2-x^2}{2yz}=\frac{z(y-x)+z^2}{2yz}=\frac{y-x+z}{2y}=\frac{y-x+y+x}{2y}=1\)
\(\frac{x^2+z^2-y^2}{2xz}=1-(-1)-1=1\)
Ta có đpcm.
Các TH còn lại tương tự.
Vậy........
ĐK: x,y khác 0
Áp dụng BĐT Cô-si ta có:
\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}\\ \ge2\sqrt{x^2.\dfrac{1}{x^2}}+2\sqrt{y^2.\dfrac{1}{y^2}}\\ =2+2=4\)
Dấu bằng xảy ra khi và chỉ khi: \(x=y=\pm1\)
Ta có:
\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\\ \Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=0\\ \Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
Do \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\) và \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\) nên:
\(\left(x-\dfrac{1}{x}\right)^2=\left(y-\dfrac{1}{y}\right)^2=0\)
Do đó: \(x=y=\pm1\)