\(\hept{\begin{cases}x+y+z=3\left(1\right)\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\left(2\right)\\x^2+y^2+z^2=17\left(3\right)\end{cases}}\left(DK:x,y,z\ne0\right)\)

Ta co:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=3>\frac{1}{3}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>\frac{1}{3}\)

Vay HPT vo nghiem

Đặt\(x^4+x^2+1=a^2\) với \(a\in Z\)

Ta có:\(x^4+x^2+1=a^2\)

\(\Leftrightarrow\left(x^4+2x^2+1\right)-x^2=a^2\)

\(\Leftrightarrow\left(x^2+1\right)^2-x^2=a^2\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x+1\right)=a^2\)

Để \(x^4+x^2+1\) là số chính phương thì:

\(x^2-x+1=x^2+x+1\Rightarrow-x=x\Rightarrow x=0\)

Vậy với  \(x=0\) thì \(x^4+x^2+1\) là số chính phương.

 xét 2  hiệu sau

(\(3\sqrt{2}-1\))²-(2\(\sqrt{3}\))²=(19-6\(\sqrt{2}\))-12=7-6\(\sqrt{2}\)=18-(11+6\(\sqrt{2}\)) = (3\(\sqrt{2}\))²-(3+\(\sqrt{2}\))² <0

(vì \(3\sqrt{2}\)<3+\(\sqrt{2}\) <=>2\(\sqrt{2}\)<3 <=>8<9 đúng)

=>3\(\sqrt{2}-1< 2\sqrt{3}\)=>\(1-3\sqrt{2}>-2\sqrt{3}\)

Câu hỏi hơi xàm

Do a;b;c không âm \(\Rightarrow\frac{a}{a+1}\ge0\) ; \(\frac{b}{b+1}\ge0\); \(\frac{c}{c+1}\ge0\)

\(\Rightarrow T\ge0\)

\(T_{min}=0\) khi \(a=b=c=0\) 

\(DK:x>0\)

Dat \(\hept{\begin{cases}\sqrt{x-\frac{2}{x}}=a\\\sqrt{2-\frac{2}{x}}=b\end{cases}}\left(a,b\ge0\right)\)

\(\Rightarrow a^2-b^2=x-2\)

Ta co HPT:

\(\hept{\begin{cases}a+b=x\\a^2-b^2=x-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=x\\\left(a+b\right)\left(a-b\right)=x-2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=x\\a-b=1-\frac{2}{x}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=x\left(1\right)\\2a=x-\frac{2}{x}+1\end{cases}\left(2\right)}\) 

Xet PT(2)

\(2\sqrt{x-\frac{2}{x}}=x-\frac{2}{x}+1\)

Dat \(\sqrt{x-\frac{2}{x}}=t\left(t\ge0\right)\)

\(\Rightarrow2t=t^2+1\)

\(\Leftrightarrow\left(t-1\right)^2=0\)

\(\Leftrightarrow t=1\left(n\right)\)

Ta lai co:

\(t=1\)

\(\Leftrightarrow\sqrt{x-\frac{2}{x}}=1\)

\(\Leftrightarrow x^2-x-2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(l\right)\\x=2\left(n\right)\end{cases}}\)

Vay nghiem cua PT la \(x=2\)

\(2\left(x^2+y^2+z^2+xy+yz+xz\right)=\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\)

\(=\left(3-x\right)^2+\left(3-y\right)^2+\left(3-z\right)^2\)

\(=27-6\left(x+y+z\right)+x^2+y^2+z^2\)

\(=9+x^2+y^2+z^2\)

Dễ dàng CM được \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=3\)

=>\(2\left(x^2+y^2+z^2+xy+yz+zx\right)\ge12\)

=> dpcm

Ta có: \(2\left(x^2+y^2+z^2+xy+yz+xz\right)\)

\(=2x^2+2y^2+2z^2+2xy+2yz+2xz\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2+2xz+z^2\right)\)

\(=\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2\)(1)

Mà \(x+y+z=3\Rightarrow\hept{\begin{cases}x+y=3-z\\y+z=3-x\\x+z=3-y\end{cases}}\)

\(\Rightarrow\left(1\right)=\left(3-z\right)^2+\left(3-x\right)^2+\left(3-y\right)^2\)

\(=9-6z+z^2+9-6x+x^2+9-6y+y^2\)

\(=27-6\left(x+y+z\right)+x^2+y^2+z^2\)

\(=9+x^2+y^2+z^2\)

Áp dụng BĐT Cauchy cho 3 số:

\(x^2+y^2+z^2=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}\ge\frac{\left(x+y+z\right)^2}{1+1+1}=\frac{3^2}{3}=3\)

\(\Rightarrow9+x^2+y^2+z^2\ge12\)

hay \(2\left(x^2+y^2+z^2+xy+yz+xz\right)\ge12\)

\(\Leftrightarrow x^2+y^2+z^2+xy+yz+xz\ge6\left(đpcm\right)\)

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