Cho \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=\frac{3}{2}\end{matrix}\right.\). Tìm GTNN của biểu thức A= \(\frac{x}{yz}\)+\(\frac{y}{zx}\)+\(\frac{z}{xy}\)
a)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\xy+yz+zx=27\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x-y=7\\x^3+y^3=133\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2-5x+y=0\\x-\sqrt{y}+1=0\end{matrix}\right.\)
Tìm GTNN của biểu thức:
\(A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{x+z}\)
Biết\(\left\{{}\begin{matrix}x.y.z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\end{matrix}\right.\)
\(A\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{1}{2}\left(x+y+z\right)\ge\dfrac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\dfrac{1}{2}\)
\(A_{min}=\dfrac{1}{2}\) khi \(x=y=z=\dfrac{1}{3}\)
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\zy+yz+zx=27\end{matrix}\right.\)
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.
\(x+y+z=9\Rightarrow\left(x+y+z\right)^2=81=3.27=3\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\Rightarrow\frac{3}{x}=1\Rightarrow x=y=z=3\)
\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\xy+yz+zx=27\end{matrix}\right.\)
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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
cho\(\left\{{}\begin{matrix}a+b+c=1\\a^2+b^2+c^2\\\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\end{matrix}\right.=1\) . CMR: xy+yz+zx=0
Theo t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\)
\(\Leftrightarrow\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\left(x+y+z\right)^2\left(1\right)\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\Leftrightarrow\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow x^2+y^2+z^2=\left(x+y+z\right)^2\)
\(\Leftrightarrow2\left(xy+yz+xz\right)=0\Leftrightarrow xy+yz+xz=0\left(đpcm\right)\)
1.Giải hệ pt
1)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\xy+yz+zx=3\\\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=x\end{cases}}\)
2)\(\hept{\begin{cases}xy+yz+zx=3\\\left(x+y\right)\left(y+z\right)=\sqrt{3}z\left(1+y^2\right)\\\left(y+z\right)\left(z+x\right)=\sqrt{3}x\left(1+z^2\right)\end{cases}}\)
3)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
giải hệ pt
\(\left\{\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{3}{2}\\\frac{x\text{z}}{x+z}=\frac{6}{7}\end{matrix}\right.\)
Bài này đơn giản thôi :))
\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} \frac{x+y}{xy}=\frac{3}{2}\\ \frac{y+z}{yz}=\frac{2}{3}\\ \frac{x+z}{xz}=\frac{7}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{3}{2}\\ \frac{1}{y}+\frac{1}{z}=\frac{2}{3}\\ \frac{1}{x}+\frac{1}{z}=\frac{7}{6}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{2}{x}=\frac{3}{2}+\frac{7}{6}-\frac{2}{3}\\ \frac{2}{y}=\frac{3}{2}+\frac{2}{3}-\frac{7}{6}\\ \frac{2}{z}=\frac{2}{3}+\frac{7}{6}-\frac{3}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\\ z=6\end{matrix}\right.\)
Vậy $(x,y,z)=(1,2,6)$ là nghiệm của hệ phương trình
1. Giải hpt: \(\left\{{}\begin{matrix}x+y+z=0\\2x+3y+z=0\\\left(x+1\right)^2+\left(y+2\right)^2+\left(z+3\right)^2=26\end{matrix}\right.\)
2. Cho x,y,z là nghiệm của hpt : \(\left\{{}\begin{matrix}\frac{x}{3}+\frac{y}{12}-\frac{z}{4}=1\\\frac{x}{10}+\frac{y}{5}+\frac{z}{3}=1\end{matrix}\right.\) . Tính \(A=x+y+z\)
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):
\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)