Ta có:
\(\left(1.x+3.y+4.z+1.t\right)^2\le\left(1^2+3^2+4^2+1^2\right)\left(x^2+y^2+z^2+t^2\right)\)
\(\Leftrightarrow\left(x+3y+4z+t\right)^2\le27\left(x^2+y^2+z^2+t^2\right)\)
Dấu "=" xảy ra khi và chỉ khi: \(x=\frac{y}{3}=\frac{z}{4}=t\Leftrightarrow\left\{{}\begin{matrix}y=3x\\z=4x\\t=x\end{matrix}\right.\)
Thay vào pt dưới:
\(x^3+27x^3+64x^3+x^3=93\)
\(\Leftrightarrow x=1\Rightarrow\left\{{}\begin{matrix}y=3\\z=4\\t=1\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}x^3\left(y^2+3y+3\right)=3y^2\\y^3\left(z^2+3z+3\right)=3z^2\\z^3\left(x^2+3x+3\right)=3x^2\end{matrix}\right.\)
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 \(A=x+y+z\)
giải hệ phương trình:
a)\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\5yz=6\left(y-z\right)\\4xz=3\left(x+y\right)\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\\7x-3y+2z=37\end{matrix}\right.\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\9yz=20\left(y+z\right)\\8zx=15\left(z+x\right)\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x+y+z=3\\y+z+t=4\\z+t+x=5\\t+x+y=6\end{matrix}\right.\)
Giải hệ pt và pt sau:
a.\(\left\{{}\begin{matrix}\left(2x-3\right)\cdot\left(2y+4\right)=4x\cdot\left(y-3\right)+54\\\left(x+1\right)\cdot\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}x+y-1=0\\x^2+xy+3=0\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2x-3y=5\\x^2-y^2=40\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}3x+2y=36\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)
e.\(\left\{{}\begin{matrix}2x+y=5m-1\\x-2y=2\end{matrix}\right.\) . Tìm m để hệ có nghiệm (x;y) t/m x\(^2\)-2y\(^2\)=1
f. \(\frac{t^2}{t-1}+t=\frac{2t^2+5t}{t+1}\)
g.\(\frac{x^2+2x-3}{x^2-9}+\frac{2x^2-2}{x^2-3x+2}=8\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}3x-2y+z=14\\2x+y-z=3\\z-2x=-5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-y^2=4y+2x+3\\x^2+2x+y=0\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left|xy-4\right|=8-y^2\\xy=2+x^2\end{matrix}\right.\)
Tìm nghiệm nguyên của hệ phương trình:
\(\left\{{}\begin{matrix}x+y+z=3\left(1\right)\\x^3+y^3+z^3=3\left(2\right)\end{matrix}\right.\)
Tìm nghiệm nguyên của hệ phương trình sau: \(\left\{{}\begin{matrix}x^3-y^3-z^3=3xyz\\x^2=2\left(y+z\right)\end{matrix}\right.\)
