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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
\(\left\{{}\begin{matrix}x+y+z=6\left(1\right)\\xy+yz-zx=7\\x^2+y^2+z^2=14\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\left(x+y+z\right)^2=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}xy+yz+xz=11\\xy+yz-xz=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}xy+yz=\frac{11+7}{2}=9\\xz=\frac{11-7}{2}=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y\left(x+z\right)=9\\x=\frac{2}{z}\end{matrix}\right.\)
=>\(y\left(\frac{2}{z}+z\right)=9\)
<=> \(y=\frac{9}{\frac{2}{z}+z}=\frac{9}{\frac{2+z^2}{z}}=\frac{9z}{2+z^2}\)
Thay \(x=\frac{2}{z},y=\frac{9z}{2+z^2}\) vào (1) có:
\(\frac{2}{z}+\frac{9z}{2+z^2}+z=6\)
<=> \(\frac{2\left(2+z^2\right)+9z^2+z^2\left(2+z^2\right)}{z\left(2+z^2\right)}=6\)
<=>\(4+2z^2+9z^2+2z^2+z^4=6z\left(2+z^2\right)\)
<=> \(z^4+13z^2+4-12z-6z^3=0\)
<=> \(z^4-3z^3+2z^2-3z^3+9z^2-6z+2z^2-6z+4=0\)
<=>\(z^2\left(z^2-3z+2\right)-3z\left(z^2-3z+2\right)+2\left(z^2-3z+2\right)=0\)
<=> \(\left(z^2-3z+2\right)^2=0\)
<=> \(z^2-3z+2=0\)<=> \(z\left(z-2\right)-\left(z-2\right)=0\)
<=> \(\left(z-1\right)\left(z-2\right)=0\)
=>\(\left[{}\begin{matrix}z=1\\z=2\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\frac{2}{z}=2,y=\frac{9z}{2+z^2}=3\\x=1,y=3\end{matrix}\right.\)
Vậy (x,y,z) \(\in\left\{\left(2,3,1\right),\left(1,3,2\right)\right\}\)
1/Liên hợp đi cho nó nhẹ:D
ĐKXĐ: \(x\ge16\)
PT \(\Leftrightarrow\sqrt{x+24}-7+\sqrt{x-16}-3=0\)
\(\Leftrightarrow\frac{x-25}{\sqrt{x+24}+7}+\frac{x-25}{\sqrt{x-16}+3}=0\)
\(\Leftrightarrow\left(x-25\right)\left(\frac{1}{\sqrt{x+24}+7}+\frac{1}{\sqrt{x-16}+3}\right)=0\)
\(\Leftrightarrow x=25\)
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 \((x,y,z)=(2,3,1),(1,3,2)\)
@Nguyễn Huy Thắng @Akai Haruma @Hoàng Lê Bảo Ngọc @Trần Việt Linh @Nguyễn Huy Tú Nguyễn Phương Trâm Hung nguyen ......................