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\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
Lấy (2) cộng (3) ta được
\(x^2+y^2-yz-zx=2\) (4)
Lấy (1) - (4) ta được
\(2x\left(x+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-z\end{matrix}\right.\)
Xét 2 TH rồi thay vào tìm được y và z
1. \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{6}{5}\\\dfrac{y+z}{yz}=\dfrac{3}{2}\\\dfrac{z+x}{zx}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{5}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{7}{10}\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
Lời giải:
\(\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=36\)
Kết hợp với \(x^2+y^2+z^2=14\Rightarrow xy+yz+xz=11\)
Có \(\left\{\begin{matrix} xy+yz-xz=7\\ xy+yz+xz=11\end{matrix}\right.\Rightarrow \left\{\begin{matrix} xz=2\\ xy+yz=9\rightarrow y(6-x)=9\rightarrow y=3\rightarrow x+z=3\end{matrix}\right.\)
Từ \(\left\{\begin{matrix} xz=2\\ x+z=3\end{matrix}\right.\Rightarrow \left[ \begin{array}{ll} (x,z)=(2,1) \\ \\ (x,z)=(1,2) \end{array} \right.\)
Vậy HPT có nghiệm \((x,y,z)=(2,3,1),(1,3,2)\)
1/Liên hợp đi cho nó nhẹ:D
ĐKXĐ: \(x\ge16\)
PT \(\Leftrightarrow\sqrt{x+24}-7+\sqrt{x-16}-3=0\)
\(\Leftrightarrow\frac{x-25}{\sqrt{x+24}+7}+\frac{x-25}{\sqrt{x-16}+3}=0\)
\(\Leftrightarrow\left(x-25\right)\left(\frac{1}{\sqrt{x+24}+7}+\frac{1}{\sqrt{x-16}+3}\right)=0\)
\(\Leftrightarrow x=25\)
\(\left\{{}\begin{matrix}xy=x+y+1\\yz=y+z+5\\zx=z+x+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-x-y-1=-2\\yz-y-z-1=4\\zx-z-x-1=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=-2\\\left(y-1\right)\left(z-1\right)=4\\\left(z-1\right)\left(x-1\right)=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy+y+x+1=10\\yz+y+z+1=5\\zx+x+z+1=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=10\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=2\end{matrix}\right.\)
\(\Rightarrow\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=100\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=10\)
\(\Rightarrow\left\{{}\begin{matrix}z+1=1\\x+1=2\\y+1=5\end{matrix}\right.\)
\(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\Rightarrow x-y=y-z=z-x=0\)\(\Rightarrow x=y=z\)
\(\Rightarrow x^{2010}+y^{2010}+z^{2010}=3x^{2010}=3^{2010}\)
\(\Rightarrow x^{2010}=\dfrac{3^{2010}}{3}=3^{2009}\Rightarrow x=\sqrt[2010]{3^{2009}}\)
\(\Rightarrow x=y=z=\sqrt[2010]{3^{2009}}\)
Lời giải:
PT (1)
\(\Leftrightarrow x^2+y^2+z^2-(xy+yz+xz)=0\)
\(\Leftrightarrow 2(x^2+y^2+z^2)-2(xy+yz+xz)=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Thấy rằng \((x-y)^2; (y-z)^2; (z-x)^2\geq 0\forall x,y,z\in\mathbb{R}\)
\(\Rightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} (x-y)^2=0\\ (y-z)^2=0\\ (z-x)^2=0\end{matrix}\right.\Leftrightarrow x=y=z\)
Thay vào PT (2)
\(\Leftrightarrow x^{2010}+x^{2010}+x^{2010}=3^{2010}\)
\(\Leftrightarrow 3.x^{2010}=3^{2010}\Leftrightarrow x^{2010}=3^{2009}\)
\(\Leftrightarrow x=\sqrt[2010]{3^{2009}}\)
Vậy \((x,y,z)=(\sqrt[2010]{3^{2009}},\sqrt[2010]{3^{2009}},\sqrt[2010]{3^{2009}})\)