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NV
18 tháng 10 2020

ĐKXĐ: ...

\(\left\{{}\begin{matrix}x+y+z=3\\xy+yz+zx=\frac{1}{3}xyz\\x^2+y^2+z^2=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=3\\xy+yz+zx=\frac{1}{3}xyz\\xy+yz+zx=\frac{\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)}{2}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=3\\xy+yz+zx=-4\\xyz=-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+z=3-x\\yz=-4-x\left(y+z\right)=-4-x\left(3-x\right)\\xyz=-12\end{matrix}\right.\)

\(\Rightarrow x\left(-4-3x+x^2\right)=-12\)

\(\Leftrightarrow x^3-3x^2-4x+12=0\)

\(\Leftrightarrow...\)

NV
11 tháng 2 2020

a/ Đơn giản là dùng phép thế:

\(x+2y+x+y+z=0\Rightarrow x+2y=0\Rightarrow x=-2y\)

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

Thế vào pt cuối:

\(\left(1-2y\right)^2+\left(y+2\right)^2+\left(y+3\right)^2=26\)

Vậy là xong

b/ Sử dụng hệ số bất định:

\(\left\{{}\begin{matrix}a\left(\frac{x}{3}+\frac{y}{12}-\frac{z}{4}\right)=a\\b\left(\frac{x}{10}+\frac{y}{5}+\frac{z}{3}\right)=b\end{matrix}\right.\)

\(\Rightarrow\left(\frac{a}{3}+\frac{b}{10}\right)x+\left(\frac{a}{12}+\frac{b}{5}\right)y+\left(\frac{-a}{4}+\frac{b}{3}\right)z=a+b\) (1)

Ta cần a;b sao cho \(\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}=-\frac{a}{4}+\frac{b}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}\\\frac{a}{3}+\frac{b}{10}=-\frac{a}{4}+\frac{b}{3}\end{matrix}\right.\) \(\Rightarrow\frac{a}{2}=\frac{b}{5}\)

Chọn \(\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\) thay vào (1):

\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)

22 tháng 8 2019

\(a)DK:z\ne1\)

\(\left\{{}\begin{matrix}\frac{4}{z-1}+2x=7\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{z-1}+x=\frac{7}{2}=3,5\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-5y=-5\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=-8\\5x-3y=3\\\frac{2}{z-1}+y=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\5x=15\\\frac{2}{z-1}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\left(T/m\right)\)

Vậy ...

\(b)DK:\left\{{}\begin{matrix}x,y,z\ne0\\x,y,z>0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+\frac{1}{y}=2\\y+\frac{1}{z}=2\\z+\frac{1}{x}=2\end{matrix}\right.\)

\(\Leftrightarrow x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}=6\)

\(\Leftrightarrow\left(x-2.\sqrt{x}.\frac{1}{\sqrt{x}}+\frac{1}{x}\right)+\left(y-2.\sqrt{y}.\frac{1}{\sqrt{y}}+\frac{1}{y}\right)+\left(z-2\sqrt{z}.\frac{1}{\sqrt{z}}+\frac{1}{z}\right)+2+2+2=6\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=0\)

\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2;\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2;\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=\frac{1}{\sqrt{x}}\\\sqrt{y}=\frac{1}{\sqrt{y}}\\\sqrt{z}=\frac{1}{\sqrt{z}}\end{matrix}\right.\)

\(\Leftrightarrow x=y=z=1\left(T/m\right)\)

Vậy ...

giải hệ phương trình 1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\) 2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\) 3 ,...
Đọc tiếp

giải hệ phương trình

1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)

3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)

4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)

5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)

6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)

7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)

8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)

9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

3
NV
4 tháng 8 2020

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x+y+z}{x\left(y+z\right)}=\frac{1}{2}\\\frac{x+y+z}{y\left(z+x\right)}=\frac{1}{3}\\\frac{x+y+z}{z\left(x+y\right)}=\frac{1}{4}\end{matrix}\right.\) lần lượt chia vế cho vế ta được hệ:

\(\left\{{}\begin{matrix}\frac{y\left(z+x\right)}{x\left(y+z\right)}=\frac{3}{2}\\\frac{z\left(x+y\right)}{x\left(y+z\right)}=2\\\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2yz=xy+3zx\\yz=2xy+xz\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2yz=xy+3zx\\3yz=6xy+3zx\end{matrix}\right.\)

\(\Rightarrow yz=5xy\Rightarrow z=5x\)

Thế vào \(yz=2xy+zx\Rightarrow5xy=2xy+5x^2\)

\(\Leftrightarrow3xy=5x^2\Rightarrow y=\frac{5x}{3}\)

Thế vào pt đầu: \(\frac{1}{x}+\frac{1}{\frac{5x}{3}+5x}=\frac{1}{2}\Rightarrow\frac{23}{20x}=\frac{1}{2}\Rightarrow x=\frac{23}{10}\)

\(\Rightarrow y=\frac{23}{6};z=\frac{23}{2}\)

4 tháng 8 2020

Dạ em cảm ơn ạ.

NV
5 tháng 10 2019

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

\(\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

Thay vào pt đầu và cuối