a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x< -2\end{matrix}\right.\)

b) ĐKXĐ: \(-\sqrt{2}\le x\le\sqrt{2}\)

c) ĐKXĐ: \(x\ge1\)

x∈[0, ∞)

a) Biểu thức xác định `<=> x^2-2x-1>0`

`<=>(x^2-2x+1)-2>0`

`<=>(x-1)^2-(\sqrt2)^2>0`

`<=>(x-1+\sqrt2)(x-1-\sqrt2)>0`

`<=>` \(\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\)

`D=(-∞; 1-\sqrt2) \cup (1+\sqrt2 ; +∞)`

b) Biểu thức xác định `<=> x-\sqrt(2x+1)>0`

`<=> x>\sqrt(2x+1)`

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x+1\ge0\\x^2>2x+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x< 1-\sqrt{2}\\x>1+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow x>1+\sqrt{2}\)

`D=(1+\sqrt2 ; +∞)`

ĐKXĐ: 2x-1>=0 và \(x-\sqrt{2x-1}>0\)

=>x>=1/2 và x>căn 2x-1

=>x>=1/2 và x^2>2x-1

=>x>=1/2 và x^2-2x+1>0

=>x>=1/2 và x<>1

\(\dfrac{1}{\sqrt[]{x-\sqrt[]{2x-1}}}\left(1\right)\)

\(\left(1\right)xđ\Leftrightarrow\left\{{}\begin{matrix}x-\sqrt[]{2x-1}>0\\2x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt[]{2x-1}< x\left(2\right)\\x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\left(I\right)\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x-1\ge0\\2x-1< x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge\dfrac{1}{2}\\x^2+2x-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge\dfrac{1}{2}\\\left(x-1\right)^2>0,\forall x\ne0\end{matrix}\right.\)  \(\Leftrightarrow x\ge\dfrac{1}{2}\)

\(\left(I\right)\Leftrightarrow x\ge\dfrac{1}{2}\)

Đính chính

\(...\left(x-1\right)^2>0,\forall x\ne1\)

\(...\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\end{matrix}\right.\)

ĐKXĐ \(a\ge0,a\ne1\)

Ta có: \(\sqrt{6-4\sqrt{2}}=\sqrt{\left(2-\sqrt{2}\right)^2}=2-\sqrt{2}\)

        \(\sqrt[3]{20+14\sqrt{2}}=\sqrt[3]{2\sqrt{2}+12\sqrt{2}+8+12}=\sqrt[3]{\left(\sqrt{2}+2\right)^3}=2+\sqrt{2}\)

          \(\sqrt[3]{\left(a+3\right)\sqrt{a}-3a-1}=\sqrt[3]{\left(\sqrt{a}-1\right)^3}=\sqrt{a}-1\)

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

 Khi đó \(P=\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)+\sqrt{a}-1.\frac{2}{\sqrt{a}-1}\)

               \(=2+2=4\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)

\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}\)

\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)

c) Để M=-4 thì \(-2\sqrt{a}=-4\)

\(\Leftrightarrow\sqrt{a}=2\)

hay a=4(thỏa ĐK)

ĐKXĐ: \(x-\sqrt{\left(x-2\right)^2}>=0\)

=>x>=|x-2|

=>x^2>=(x-2)^2 và x>=0

=>0>=-4x+4 và x>=0

=>x>=0 và -4x+4<=0

=>x>=0 và -4x<=-4

=>x>=1

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

\(DKXD:\left[{}\begin{matrix}x-x+2\ge0\\x+x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2\ge0\left(LD\right)\\2x\ge2\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x-1\ge0\\x+2>0\\3-x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x>-2\\x\le3\end{matrix}\right.\) 

\(\Rightarrow\dfrac{1}{2}\le x\le3\)

a) ĐKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)

b) Với \(a>0;a\ne1;a\ne4\), ta có:

\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ =\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

c)\(B\le\dfrac{1}{3}\rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\rightarrow\dfrac{-2}{\sqrt{a}}\le0\) (đúng với mọi a thoả ĐKXĐ).