Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\Rightarrow\hept{\begin{cases}a+b+c=1\\a;b;c>0\end{cases}}\)

Và \(\frac{ab}{\sqrt{a^2+b^2+2c^2}}+\frac{bc}{\sqrt{b^2+c^2+2a^2}}+\frac{ca}{\sqrt{c^2+a^2+2b^2}}\le\frac{1}{2}\)

Ta có :

\(\frac{ab}{a^2+b^2+2c^2}=\frac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)

\(\le\frac{2ab}{a+b+2c}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

Tương tự cho 2 BĐT còn lại roouf cộng theo vế :

\(VT\le\frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{1}{2}\)

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\Rightarrow x=y=z=\frac{1}{9}\)

Chúc bạn học tốt !!!

bn mũ 3 lên đc bao nhiêu đã

sau đó p/t thành nhân tử đặt nhân tử chung

hok tốt

PT <=> \(x+5+x+6=2x+11\)

\(2x+11=2x+11\Leftrightarrow0=0\)

Áp dụng BĐT Mincopxki và AM - GM ta có :

\(P=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\)

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

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{9}{x+y+z}\right)^2}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}+\frac{80}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{\sqrt[2]{\left(x+y+z\right)^2.\frac{1}{\left(x+y+z\right)^2}+80}}\)

\(\ge\sqrt{2+80}=\sqrt{82}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

Chúc bạn học tốt !!!

Câu này em tui đăng chứ đâu biết đâu

IQ vô cực thì tự làm đi

\(DK:x\in\left[-2;2\right]\)

\(\Leftrightarrow\left(2-x\right)+\sqrt{4-x^2}\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2-x\right)+\sqrt{\left(2+x\right)\left(2-x\right)}\left(3x-1\right)=0\)

\(\Leftrightarrow\sqrt{2-x}\left[\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(1\right)\\\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)=0\left(2\right)\end{cases}}\)

Xet PT(2)

\(\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2+x\right)\left(3x-1\right)^2=2-x\)

\(\Leftrightarrow\left(2+x\right)\left(9x^2-6x+1\right)=2-x\)

\(\Leftrightarrow9x^3+12x^2-10x=0\)

\(\Leftrightarrow x\left(9x^2+12x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(n\right)\\9x^2+12x-10=0\end{cases}}\)

Xet PT 

\(9x^2+12x-10=0\)

Ta co:

\(\Delta^`=6^2-9.\left(-10\right)=126>0\)

\(\Rightarrow\hept{\begin{cases}x_1=\frac{-4+\sqrt{14}}{6}\left(n\right)\\x_2=\frac{-4-\sqrt{14}}{6}\left(n\right)\end{cases}}\)

Vay tap nghiem cua PT la \(S=\left\{\frac{-4-\sqrt{14}}{6};\frac{-4+\sqrt{14}}{6};0;2\right\}\)