\(\left\{{}\begin{matrix}x+y+z=6\left(1\right)\\xy+yz-zx=7\\x^2+y^2+z^2=14\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\left(x+y+z\right)^2=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\\x^2+y^2+z^2=14\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14+2\left(xy+yz+xz\right)=36\\xy+yz-xz=7\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}xy+yz+xz=11\\xy+yz-xz=7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}xy+yz=\frac{11+7}{2}=9\\xz=\frac{11-7}{2}=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y\left(x+z\right)=9\\x=\frac{2}{z}\end{matrix}\right.\)

=>\(y\left(\frac{2}{z}+z\right)=9\)

<=> \(y=\frac{9}{\frac{2}{z}+z}=\frac{9}{\frac{2+z^2}{z}}=\frac{9z}{2+z^2}\)

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) \(\in\left\{\left(2,3,1\right),\left(1,3,2\right)\right\}\)

a.

\(\frac{3x-36}{12}=\frac{5y-45}{15}=\frac{z-1}{1}=\frac{3x+5y-z-50}{26}=\frac{-48}{26}\)

\(\Rightarrow\frac{x-12}{4}=\frac{-48}{26}\Rightarrow x=...\)

Tương tự với y, z, nhưng chắc bạn nhầm đề, nếu pt bên dưới là -2 thì nó ra \(\frac{-52}{26}=-2\) kết quả đẹp hơn nhiều

b. Không rõ đề

c.

\(x+y+z=9\Rightarrow\left(x+y+z\right)^2=81=3.27=3\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\Rightarrow\frac{3}{x}=1\Rightarrow x=y=z=3\)

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 ......................

\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)

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

2. \(\left\{{}\begin{matrix}\left(xy-x\right)-\left(y-1\right)=6\\\left(yz-y\right)-\left(z-1\right)=12\\\left(zx-z\right)-\left(x-1\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=6\\\left(y-1\right)\left(z-1\right)=12\\\left(z-1\right)\left(x-1\right)=8\end{matrix}\right.\)

Đến đây dễ rồi

1. Với mọi số thực x;y;z ta có:

\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)

\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)

\(\Rightarrow P\ge3\)

\(P_{min}=3\) khi \(x=y=z=1\)

1.1

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)

\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)

\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)

\(\Leftrightarrow a=b\Leftrightarrow x=y\)

Thay vào pt đầu:

\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))

\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)

\(\Rightarrow a=1\Rightarrow x=y=1\)

2.

\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)

\(\Rightarrow4x^2-10xy+4y^2=0\)

\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)

Thế vào pt đầu

...

x + y + z = 6 => (x + y + z)² = 36

=> x² + y² + z² + 2(xy + yz + zx) = 36

=> x² + y² + z² = 36 - 2.12 = 12

=> x² + y² + z² = xy + yz + zx

Ta có VT \(\ge\) VP. Dấu "=" xảy ra <=> x = y = z

Thay vào hệ ta có (x; y; z) = (2; 2; 2)

Lời giải:

Ta có: \(\left\{\begin{matrix} x^2-|x|=|yz|\\ y^2-|y|=|xz|\end{matrix}\right.\Rightarrow x^2-y^2-(|x|-|y|)=|yz|-|zx|\)

\(\Leftrightarrow (|x|-|y|)(|x|+|y|)-(|x|-|y|)=|z|(|y|-|x|)\)

\(\Leftrightarrow (|x|-|y|)(|x|+|y|-1+|z|)=0\)

Từ đây xét các TH:

TH1: \(|x|-|y|=0\Leftrightarrow |x|=|y|\)

Thay vào pt đầu tiên: \(x^2-|x|=|yz|=|xz|\)

\(\Leftrightarrow |x|(|x|-1-|z|)=0\)

\(\Leftrightarrow \left[\begin{matrix} |x|=0\\ |x|-1-|z|=0\end{matrix}\right.\)

+) Với \(|x|=0\Rightarrow x=0\rightarrow y=0\).

Thay vào PT(3): \(z^2-|z|=0\Leftrightarrow z=0; z=\pm 1\)

+) Với \(|x|-1-|z|=0\Leftrightarrow |y|=|x|=|z|+1\)

Thay vào PT(3): \(z^2-|z|=(|z|+1)^2=z^2+1+2|z|\)

\(\Leftrightarrow 1+3|z|=0\) (vô lý)

TH2: \(|x|+|y|+|z|=1\)

\(\Rightarrow |x|-1=-(|y|+|z|)\leq 0\)

Khi đó xét PT(1): \(|yz|=x^2-|x|=|x|(|x|-1)\) ta thấy:

VP luôn nhỏ hơn hoặc bằng $0$

Mà vế trái luôn lớn hơn hoặc bằng $0$. Do đó để hai vế bằng nhau thì:

\(|yz|=|x|(|x|-1)=0\). Kết hợp với \(|x|+|y|+|z|=1\)

Từ đây ta dễ dàng thu được

\((x,y,z)=(0,0,\pm 1), (\pm 1, 0,0), (0,\pm 1, 0)\)