a: ĐKXĐ: x^2-1>=0 

=>x>=1 hoặc x<=-1

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

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

x>=căn 2

=>x^2>=2

=>x^2-1>=1

=>căn x^2-1>=1

=>căn(x^2-1)-1>=0

=>\(A=\sqrt{x^2-1}+1-\sqrt{x^2+1}+1=2\)

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

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

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

a) A có nghĩa <=> \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\)

b) Nếu \(x\ge\sqrt{2}\)khi đó \(\sqrt{x^2-1}-1\ge\sqrt{\left(\sqrt{2}\right)^2-1}-1=0\)

Ta có: \(A=\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=2\)

a) ĐK; x>1; x<-1

b)\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)\(=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)

Nếu \(x\ge\sqrt{2}\Rightarrow x^2\ge2\Leftrightarrow x^2-1\ge1\Leftrightarrow\sqrt{x^2-1}\ge1\Leftrightarrow\sqrt{x^2-1}-1\ge0\Rightarrow\left|\sqrt{x^2-1}-1\right|=\sqrt{x^2-1}-1\)

\(\Leftrightarrow A=\sqrt{x^2-1}+1-\sqrt{x^2-1}+1=2\)

Đúng nha

Ta có: 

\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)

\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)

\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)

\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2}{2}\)

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

\(P=\sqrt{x}-x\)

b) Để \(P>0\) thì \(\sqrt{x}-x>0\)

   \(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Suy ra: TH₁: \(\sqrt{x}< 0\) và \(1-\sqrt{x}< 0\) (Loại) vì \(\sqrt{x}\ge0\)

            TH₂:\(\sqrt{x}>0\)  và \(1-\sqrt{x}>0\) (Nhận)

Ta có  \(\Rightarrow x>0\) \(\Rightarrow\sqrt{x}< 1\) \(\Rightarrow x< 1\)

Vậy để \(P>0\) thì \(0< x< 1\)

c)\(P=\sqrt{x}-x\)

\(P=-\left(x-\sqrt{x}\right)\)

\(P=-\left(\left(\sqrt{x}\right)^2-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)

\(P=-\left(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\right)\)

\(P=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)

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

\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)

Nên \(-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) \(\Rightarrow x=\frac{1}{4}\)

Vậy GTLN của \(P\) là \(\frac{1}{4}\) khi \(x=\frac{1}{4}\)

a. 2

@Doraemon2611.

a) ta có : \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)

\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)

\(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{1-x}{\sqrt{2}}\right)^2\)

b) ta có : \(x< 1\Leftrightarrow x-1< 0\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x-\sqrt{x}< 0\Leftrightarrow\sqrt{x}-x>0\)

\(\Leftrightarrow P>0\left(đpcm\right)\)

VT=|căn(x-2)+1|+|căn (x-2)-1|

=|căn (x-2)+1|+|1-căn x-2|>=|căn(x-2)+1+1-căn(x-2)|=2