a) \(ĐKXĐ:x\ge-1\)

\(\sqrt{x+1}=2\)\(\Rightarrow\left(\sqrt{x+1}\right)^2=4\)

\(\Rightarrow x+1=4\)\(\Leftrightarrow x=3\)( thỏa mãn ĐKXĐ )

Vậy \(x=3\)

b) \(ĐKXĐ:x\ge2\)

\(2\sqrt{x-2}< 6\)\(\Leftrightarrow\sqrt{x-2}< 3\)

Vì \(\sqrt{x-2}\ge0\); \(3>0\)

\(\Rightarrow\left(\sqrt{x-2}\right)^2< 9\)\(\Leftrightarrow x-2< 9\)

\(\Leftrightarrow x< 11\)

Kết hợp với ĐKXĐ \(\Rightarrow2\le x< 11\)

Vậy \(2\le x< 11\)

c) \(ĐKXĐ:x\ge4\)

 \(\sqrt{x^2-16}=-\sqrt{x-4}\)

\(\Leftrightarrow\sqrt{x^2-16}+\sqrt{x-4}=0\)

\(\Leftrightarrow\sqrt{\left(x-4\right)\left(x+4\right)}+\sqrt{x-4}=0\)

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

Vì \(\sqrt{x+4}>0\)\(\Rightarrow\sqrt{x+4}+1>0\)

\(\Rightarrow\sqrt{x-4}=0\)\(\Leftrightarrow x-4=0\)\(\Leftrightarrow x=4\)

Vậy \(x=4\)

Ta có:

\(P=5x+4y+\frac{8}{x}+\frac{9}{y}\)

\(P=\left(\frac{8}{x}+2x\right)+\left(\frac{9}{y}+y\right)+3\left(x+y\right)\)

Áp dụng BĐT Cauchy ta được:

\(P\ge2\sqrt{\frac{8}{x}\cdot2x}+2\sqrt{\frac{9}{y}\cdot y}+3\cdot5\)

\(=2\cdot4+2\cdot3+15=29\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Vậy Min(P) = 29 khi \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Cảm ơn ạ

Áp dụng BĐT Bunyakovsky ta được:

\(\left(x+y\right)\left(\frac{2020}{x}+\frac{1}{2020y}\right)\ge\left(\sqrt{x}\cdot\sqrt{\frac{2020}{x}}+\sqrt{y}\cdot\sqrt{\frac{1}{2020y}}\right)\)

\(=\left(\sqrt{2020}+\sqrt{\frac{1}{2020}}\right)^2=2020+\frac{1}{2020}+2=2022\frac{1}{2020}\)

\(\Leftrightarrow\frac{2021}{2020}\cdot S\ge2022\frac{1}{2020}\)

\(\Rightarrow S\ge2022\frac{1}{2020}\div\frac{2021}{2020}=2021\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{\sqrt{x}}{\sqrt{\frac{2020}{x}}}=\frac{\sqrt{y}}{\sqrt{\frac{1}{2020y}}}\\x+y=\frac{2021}{2020}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020y\\x+y=\frac{2021}{2020}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

Vậy Min(S) = 2021 khi \(\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

Ta có: 

Vì \(\frac{2}{3}< x< \frac{13}{2}\Rightarrow\hept{\begin{cases}3x-2>0\\10-x>0\\13-2x>0\end{cases}}\)

Khi đó: \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\)

\(=\frac{1}{3x-2}+\frac{1}{10-x}+\frac{1}{13-2x}\) \(\left(1\right)\)

Áp dụng BĐT Cauchy Schwarz ta được:

\(\left(1\right)\ge\frac{\left(1+1+1\right)^2}{3x-2+10-x+13-2x}\)

\(=\frac{3^2}{21}=\frac{3}{7}\)

Vậy với \(\frac{2}{3}< x< \frac{13}{2}\) thì \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\ge\frac{3}{7}\)

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

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

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

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

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

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