Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1

Ta có: 

\(B=x^2+4-x+\frac{1}{x^2-x+1}\)

\(B=\left(x^2-x+1+\frac{1}{x^2-x+1}\right)+3\)

Áp dụng BĐT Cauchy ta được:

\(B\ge2\sqrt{\left(x^2-x+1\right)\cdot\frac{1}{x^2-x+1}}+3=2\cdot1+3=5\)

Dấu "=" xảy ra khi: \(x^2-x+1=\frac{1}{x^2-x+1}\)

\(\Leftrightarrow\left(x^2-x+1\right)^2=1\) mà \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)

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

Vậy Min(B) = 5 khi \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

a) ĐK: \(x\ge1\)

\(\sqrt{x}-\sqrt{x+1}+\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{\sqrt{x^3-x}}{\sqrt{x-1}}\)

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

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

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

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

\(=\left(\sqrt{x-1}-1\right)^2\)

b) \(P=1\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=1\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}-1=1\\\sqrt{x-1}-1=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy x=5,x=1

\(ĐKXĐ:\hept{\begin{cases}a\ge0\\a\ne4\end{cases}}\)

\(\left(\frac{\sqrt{a}-2}{\sqrt{a}+2}-\frac{\sqrt{a}+2}{\sqrt{a}-2}\right):\frac{1}{a-4}\)

\(=\left[\frac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right].\left(a-4\right)\)

\(=\frac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)\)

\(=\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2\)

\(=\left(a-4\sqrt{a}+4\right)-\left(a+4\sqrt{a}+4\right)\)

\(=a-4\sqrt{a}+4-a-4\sqrt{a}-4=-8\sqrt{a}\)

ĐK : \(\hept{\begin{cases}a\ge0\\a\ne4\end{cases}}\)

\(=\left(\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right)\div\frac{1}{a-4}\)

\(=\left(\frac{a-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{a+4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right)\div\frac{1}{a-4}\)

\(=\left(\frac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right)\div\frac{1}{a-4}\)

\(=\frac{-8\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\times\frac{a-4}{1}\)

\(=\frac{-8\sqrt{a}}{a-4}\times\frac{a-4}{1}=-8\sqrt{a}\)

to bit ne no la 93 do ban tinh thu xem dung ko roi cho minh bit nha ok ban 

Không viết lại đề

\(B=\left|2-x\right|+\left|3-y\right|+\left|x+y-2020\right|\ge\left|2-x+3-y+x+y-2020\right|=\left|-2015\right|=2015\)Còn lại tự làm nốt

Uầy, Bunyakovsky phát ra luôn nè :))

Ta có:

 \(\left(x+3y+4z+t\right)^2\le\left(1^2+3^2+4^2+1^2\right)\left(x^2+y^2+z^2+t^2\right)=27\left(x^2+y^2+z^2+t^2\right)\)

Dấu "=" xảy ra khi: \(x=\frac{y}{3}=\frac{z}{4}=t\)

Đặt \(x=\frac{y}{3}=\frac{z}{4}=t=k\left(k\inℝ\right)\)

\(\Rightarrow\hept{\begin{cases}x=t=k\\y=3k\\z=4k\end{cases}}\) thay vào ta được: \(k^3+27k^3+64k^3+k^3=93\)

\(\Leftrightarrow93k^3=93\Rightarrow k^3=1\Rightarrow k=1\)

\(\Rightarrow\hept{\begin{cases}x=t=1\\y=3\\z=4\end{cases}}\)