giải hệ phương trình : \(\left\{{}\begin{matrix}x+y+z=9\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\\xy+yz+zx=27\end{matrix}\right.\)
Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}x+y+z=9\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\\xy+yz+xz=27\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=\dfrac{9}{9}=1\)
Dau bang xay ra khi x=y=z=3 ( vi x+y+z=9)
a) Cho x,y,z thỏa mãn x+y+z+xy+yz+zx=6. Tìm Min \(P=x^2+y^2+z^2\)
giải hệ pt : 1) \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\sqrt{2-\dfrac{1}{y}}=2\\\dfrac{1}{\sqrt{y}}+\sqrt{2-\dfrac{1}{x}}=2\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)
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
...
Giải hệ phương trình:
a) \(\left\{{}\begin{matrix}x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=5\\\left(xy-1\right)^2=x^2-y^2+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}=9\\x+y+z\le4\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y+z=3\\x^4+y^4+z^4=3xyz\end{matrix}\right.\)
b) Áp dụng bđt Svac-xơ:
\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)
=> hpt vô nghiệm
c) Ở đây x,y,z là các số thực dương
Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{3}=1\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\dfrac{xy}{x+y}=2\\\dfrac{yz}{y+z}=4\\\dfrac{zx}{z+x}=3\end{matrix}\right.\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{1}{2}\\\dfrac{y+z}{yz}=\dfrac{1}{4}\\\dfrac{z+x}{xz}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\) ( đk : x , y , z # 0 )
Cộng từng vế của các pt lại với nhau , ta có :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{12}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{13}{24}-\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{24}-\dfrac{1}{4}=\dfrac{7}{24}\)
\(\Leftrightarrow x=\dfrac{24}{7}\left(tm\right)\)
\(\Rightarrow y=\dfrac{24}{5}\left(tm\right);z=8\left(tm\right)\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
tham khảo
https://hoc24.vn/hoi-dap/tim-kiem?id=165107&q=1%2Fx%201%2F%28y%20z%29%3D1%2F3%20%201%2Fy%201%28z%20x%29%3D1%2F4%20%201%2Fz%201%2F%28x%20y%29%3D1%2F5%20%20gi%E1%BA%A3i%20h%E1%BB%87%20ph%C6%B0%C6%A1ng%20tr%C3%ACnh%20%E1%BA%A1%20m%E1%BB%8Di%20ng%C6%B0%E1%BB%9Di%20gi%E1%BA%A3i%20d%C3%B9m%20em%20v%E1%BB%9Bi%20%E1%BA%A1#:~:text=2020%20l%C3%BAc%2013%3A53-,%E2%87%94,2,-%E2%87%92y%3D23
Giải hệ phương trình sau: \(\left\{{}\begin{matrix}\dfrac{xy}{y+x}=\dfrac{12}{5}\\\dfrac{yz}{y+z}=\dfrac{18}{5}\\\dfrac{zx}{z+x}=\dfrac{36}{13}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{xy}{x+y}=\dfrac{12}{5}\\\dfrac{yz}{y+z}=\dfrac{18}{5}\\\dfrac{zx}{z+x}=\dfrac{36}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{5}{12}\\\dfrac{y+z}{yz}=\dfrac{5}{18}\\\dfrac{z+x}{zx}=\dfrac{13}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{12}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{18}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{13}{36}\end{matrix}\right.\)
Cộng vế theo vế ta thu được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{19}{18}\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{19}{36}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{4}\\\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{z}=\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=9\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(4;6;9\right)\)
Giải hệ phương trình : \(\left\{{}\begin{matrix}\dfrac{xy}{x+y}=\dfrac{2}{3}\\\dfrac{yz}{y+z}=\dfrac{6}{5}\\\dfrac{zx}{z+x}=\dfrac{3}{4}\end{matrix}\right.\).
giải hệ phương trình
\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{8}\\\dfrac{y+x}{yz}=\dfrac{3}{4}\\\dfrac{x+z}{xz}=\dfrac{5}{6}\end{matrix}\right.\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix}
\frac{1}{x}+\frac{1}{y}=\frac{3}{8}\\
\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\
\frac{1}{z}+\frac{1}{x}=\frac{5}{6}\end{matrix}\right.\Rightarrow 2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{8}+\frac{3}{4}+\frac{5}{6}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{47}{48}\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z}=\frac{47}{48}-\frac{3}{8}\\ \frac{1}{x}=\frac{47}{48}-\frac{3}{4}\\ \frac{1}{y}=\frac{47}{48}-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{48}{29}\\ y=\frac{48}{11}\\ z=\frac{48}{7}\end{matrix}\right.\)
Dùng bđt cosi để giải hệ phương trình :\(\left\{{}\begin{matrix}x,y,z\ge0\\x^3=y^2+z+2\\\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=9\end{matrix}\right.\)
Lời giải:
$x,y,z>0$ thì $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ mới xác định.
Áp dụng BĐT AM-GM:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9$
Dấu "=" xảy ra khi $x=y=z$. Thay vào pt $(2)$:
$x^3=x^2+x+2$
$\Leftrightarrow x^3-x^2-x-2=0$
$\Leftrightarrow x^2(x-2)+x(x-2)+(x-2)=0$
$\Leftrightarrow (x^2+x+1)(x-2)=0$
Dễ thấy $x^2+x+1>0$ với mọi $x>0$ nên $x-2=0$
$\Rightarrow x=2$
Vậy hpt có nghiệm $(x,y,z)=(2,2,2)$