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\(x^2+3x+2=0\)

\(\Rightarrow x^2+2x+x+2=0\)

\(\Rightarrow x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)

\(x^3-3x^2-x+3=0\)

\(\Rightarrow x^2\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x^2-1\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=3\end{matrix}\right.\)

\(x^3+4x^2+5x^2=0\)

\(\Rightarrow x^3+9x^2=0\)

\(\Rightarrow x^2\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

Khi đó:

\(\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\\dfrac{1}{x+y+z}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\2x+2y+2z=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y+2z+2=4x\\2x+2z+4=4y\\2x+2y-6=4z\\2x+2y+2z=1\end{matrix}\right.\)

\(\circledast\) Xét \(2x+2y+2z=1\Leftrightarrow2y+2z=1-2x\Leftrightarrow1-2x+2=4x\)

\(\Leftrightarrow3-2x=4x\Leftrightarrow6x=3\Leftrightarrow x=\dfrac{1}{2}\)

\(\circledast\)Xét \(2x+2y+2z=1\Leftrightarrow2x+2z=1-2y\Leftrightarrow1-2y+4=4y\)

\(\Leftrightarrow5-2y=4y\Leftrightarrow6y=5\Leftrightarrow y=\dfrac{5}{6}\)

\(\circledast\)Xét \(2x+2y+2z=1\Leftrightarrow2x+2y=1-2z\Leftrightarrow1-2z-6=4z\)

\(\Leftrightarrow-5-2z=4z\Leftrightarrow-5=6z\Leftrightarrow z=-\dfrac{5}{6}\)

Lời giải:

Ta có:1 điều luôn đúng:

\(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-2ab+b^2\ge0\\b^2-2bc+c^2\ge0\\c^2-2ac+a^2\ge0\end{matrix}\right.\)

Tương đương với:

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{matrix}\right.\)

Cộng theo 3 vế:

\(a^2+b^2+b^2+c^2+c^2+a^2\ge2ab+2bc+2ac\)

Suy ra \(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\left(đpcm\right)\)

Lời giải:

\(A=2004+\sqrt{2003-x}\)

a)Để \(A\) có nghĩa thì \(2003-x\ge0\Leftrightarrow x\le2003\)

b) Ta có:

\(A=2004+\sqrt{2003-x}=2005\)

Tương đương với:

\(\sqrt{2003-x}=1\)

Suy ra :\(\left|2003-x\right|=1\Rightarrow\left[{}\begin{matrix}2003-x=1\\2003-x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2002\\x=2004\end{matrix}\right.\)

c) Ta có:

Để \(A\) nhỏ nhất thì \(\sqrt{2003-x}\) cũng phải nhỏ nhất

\(\sqrt{2003-x}\ge0\Leftrightarrow2004+\sqrt{2003-x}\ge2004\)

Dấu "=" xảy ra khi: \(x=2003\)