\(đkxđ\Leftrightarrow\hept{\begin{cases}x-5\ge0\Rightarrow x\ge5\\7-x\ge0\Rightarrow x\le7\end{cases}\Rightarrow5\le x\le7}\)

Ta có :

\(\sqrt{x-5}+\sqrt{7-x}=2.\)

\(\Rightarrow\left(\sqrt{x-5}+\sqrt{7-x}\right)^2=2^2\)

\(\Rightarrow\sqrt{x-5}^2+2\sqrt{\left(x-5\right)\left(7-x\right)}+\sqrt{7-x}^2=4\)

\(\Rightarrow x-5+2\sqrt{\left(x-5\right)\left(7-x\right)}+7-x=4\)

\(\Rightarrow2\sqrt{\left(x-5\right)\left(7-x\right)}=2\)

\(\Rightarrow\sqrt{\left(x-5\right)\left(7-x\right)}=1\)

\(\Rightarrow\left(x-5\right)\left(7-x\right)=1\)

\(\Rightarrow-x^2+2x-35=1\)

\(\Rightarrow x^2-2x+36=0\)

\(\Rightarrow\left(x-1\right)^2+35=0\)( vô lí )

\(\Rightarrow\)Phương trình vô nghiệm

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ge3\end{cases}}\Rightarrow x\ge3\)

\(\sqrt{x^2-3x}-\sqrt{x-3}=0\)

\(\Rightarrow\sqrt{x\left(x-3\right)}-\sqrt{x-3}=0\)

\(\Rightarrow\sqrt{x}.\sqrt{x-3}-\sqrt{x-3}=0\)

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{cases}\Rightarrow\hept{\begin{cases}x-3=0\\\sqrt{x}=1\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x=1\end{cases}}}\)

\(x^2y+xy^2=0\Leftrightarrow xy\left(x+y\right)=0\Leftrightarrow\orbr{\begin{cases}xy=0\\x=-y\end{cases}}\)

TH1 \(x=-y\Rightarrow2x^2+3xy+2y^2=1\Leftrightarrow2y^2-3y^2+2y^2=1\Rightarrow y^2=1\Rightarrow y\Rightarrow x\)

TH2 \(xy=0\Rightarrow\orbr{\begin{cases}x=0\\y=0\end{cases}\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=0\\y=\frac{1}{\sqrt{2}}\end{cases}}\\\hept{\begin{cases}y=0\\\frac{1}{\sqrt{2}}\end{cases}}\end{cases}}}\)

AD BĐT X^3+Y^3>=XY(X+Y) LÀ RA

Có BĐT phụ:

\(a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Áp dụng

\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\)

\(\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{zx\left(z+x\right)+xyz}\)

\(=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}\)

\(=\frac{1}{xyz}\)

\(K\le\Sigma\sqrt{12a+\left(b+c\right)^2}=\Sigma\sqrt{12a+\left(3-a\right)^2}=\Sigma\sqrt{\left(a+3\right)^2}=12\)

dấu "=" xảy ra khi \(a=b=0;c=3\) và các hoán vị