Đề bạn sai câu b/

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

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

''='' <=> x = 4

=> Min A = \(\sqrt{2}\)và x = 4

B = |x-2011| + |x-1|

TH1: x \(\le\)1

=> B = 2012 - 2x \(\ge\)2010   ''='' <=> x = 1

TH2: 1\(\le\)x\(\le\)2011

=> B = x - 1 + 2011 - x = 2010 với mọi x t/m đkiện

TH3: x \(\ge\)2011

=> B = 2x - 2012 \(\ge\)2010 ''='' <=> x = 2011

Vậy Min B = 2010 <=> 1\(\le\)x\(\le\)2011

\(2016\sqrt{\left(x+1\right)^2}+2015\sqrt{\left(x-1\right)^2}\)

\(=2016\left|x+1\right|+2015\left|x-1\right|\) (1)

Ta thấy: \(\begin{cases}2016\left|x+1\right|\ge0\\2015\left|x-1\right|\ge0\end{cases}\)

\(\Rightarrow\left(1\right)\ge0\).Mà \(2016\left|x+1\right|+2015\left|x-1\right|\le0\)

\(\Rightarrow\begin{cases}2016\left|x+1\right|=0\\2015\left|x-1\right|=0\end{cases}\)\(\Rightarrow\begin{cases}\left|x+1\right|=0\\\left|x-1\right|=0\end{cases}\)\(\Rightarrow\begin{cases}x=-1\\x=1\end{cases}\)

Vô nghiệm (vì x ko nhận 2 giá trị khác nhau cùng lúc)

Vì \(\sqrt{\left(x+1\right)^2}\ge0;\sqrt{\left(x-1\right)^2}\ge0\)

=> \(2016.\sqrt{\left(x+1\right)^2}\ge0;2015.\sqrt{\left(x-1\right)^2}\ge0\)

=> \(2016.\sqrt{\left(x+1\right)^2}+2015.\sqrt{\left(x-1\right)^2}\ge0\)

Mà theo đề bài: \(2016.\sqrt{\left(x+1\right)^2}+2015.\sqrt{\left(x-1\right)^2}\le0\)

=> \(2016.\sqrt{\left(x+1\right)^2}+2015.\sqrt{\left(x-1\right)^2}=0\)

=> \(\begin{cases}2016.\sqrt{\left(x+1\right)^2}=0\\2015.\sqrt{\left(x-1\right)^2}=0\end{cases}\)=> \(\begin{cases}\sqrt{\left(x+1\right)^2}=0\\\sqrt{\left(x-1\right)^2}=0\end{cases}\)=> \(\begin{cases}x+1=0\\x-1=0\end{cases}\) => \(\begin{cases}x=-1\\x=1\end{cases}\)

, vô lý vì x không thể cùng lúc nhận 2 giá trị khác nhau

Vậy không tồn tại giá trị của x thỏa mãn đề bài

Ta có (x + |x| + 2016)(y + |y| + 2016) > 2016 với mọi x, y nên không thể tính được P

x+y =0

=> P = 1

x+y=0

=>P=1

a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)

\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{4x}{\sqrt{x}-3}\)

b) \(P=\dfrac{4x}{\sqrt{x}-3}\)

\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)

Theo BĐT côsi ta có:

\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)

Vậy: \(P_{min}=36\Leftrightarrow x=36\) 

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)