Tìm tất cả bộ ba số x,y,z thoả mãn:
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{3}{z}=1\\\dfrac{12}{yz}-\dfrac{1}{x^2}=1\end{matrix}\right.\)
Cho x, y, z là các số thoả mãn:
\(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{12}-\dfrac{z}{4}=1\\\dfrac{x}{10}+\dfrac{y}{5}+\dfrac{z}{3}=1\end{matrix}\right.\)
Tính \(M=x^{10}+y^{100}+z^{1000}\)
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
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Cho các số x, y, z thoả mãn: \(\left\{{}\begin{matrix}x+y+z=a\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{c}\\x^2+y^2+z^2=b^2\end{matrix}\right.\)
Tính \(P=x^3+y^3+z^3\) theo a, b, c.
Lời giải:
$xy+yz+xz=\frac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}(a^2-b^2)$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}$
$\Rightarrow xyz=c(xy+yz+xz)=\frac{1}{2}c(a^2-b^2)$
Khi đó:
$P=(x+y+z)^3-3(x+y)(y+z)(x+z)$
$=(x+y+z)^3-3[(x+y+z)(xy+yz+xz)-xyz]=(x+y+z)^3-3(xy+yz+xz)(x+y+z)+3xyz$
$=a^3-\frac{3}{2}a(a^2-b^2)+\frac{3}{2}c(a^2-b^2)$
Cho x, y, z dương thỏa mãn \(\left\{{}\begin{matrix}x^2+xy+y^2=1\\y^2+yz+z^2=\dfrac{1}{4}\\x^2+xz+z^2=\dfrac{3}{4}\end{matrix}\right.\)
Tính B=x+y+z
cho x,y,z thỏa mãn \(\left\{{}\begin{matrix}x^2+y^2+z^2=2\\xy+yz+xz=1\end{matrix}\right.\)
chứng minh \(\dfrac{-4}{3}\le x,y,z\le\dfrac{4}{3}\)
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Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}y=2\sqrt{x-1}\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}xy-\dfrac{x}{y}=9.6\\xy-\dfrac{y}{x}=7.5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
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 \(x=y=z=\dfrac{3}{3}=1\)
Cho \(\left\{{}\begin{matrix}x;y;z>=0\\x+y+z=2\end{matrix}\right.\) CMR \(\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{y^2-yz+z^2}+\dfrac{1}{z^2-xz+x^2}\ge3\)
Không mất tính tổng quát, giả sử \(x\ge y\ge z\)
\(y^2-yz+z^2=y^2+\left(z-y\right)y\le y^2\Rightarrow\dfrac{1}{y^2-yz+z^2}\ge\dfrac{1}{y^2}\)
Tương tự: \(\dfrac{1}{z^2-xz+x^2}\ge\dfrac{1}{x^2}\)
\(\Rightarrow P\ge\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{x^2-xy+y^2}+\dfrac{x^2-xy+y^2}{x^2y^2}+\dfrac{1}{xy}\)
\(P\ge2\sqrt{\dfrac{x^2-xy+y^2}{x^2y^2\left(x^2-xy+y^2\right)}}+\dfrac{1}{xy}=\dfrac{3}{xy}\ge\dfrac{12}{\left(x+y\right)^2}\ge\dfrac{12}{\left(x+y+z\right)^2}=3\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\) và hoán vị
giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z+x}=\dfrac{1}{3}\\\dfrac{1}{z}+\dfrac{1}{x+y}=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}\\x^2-y^2=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\\\dfrac{7}{\sqrt{x-7}}-\dfrac{2}{\sqrt{y+6}}=\dfrac{5}{3}\end{matrix}\right.\)