a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge6\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(-1\le x\le1\)

c: ĐKXĐ: \(x\le-2\)

a. \(\sqrt{\left(x-2\right)\left(x-6\right)}\) có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x-6\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge6\end{matrix}\right.\) \(\Leftrightarrow x\ge6\)

b. \(\sqrt{1-x^2}\) có nghĩa \(\Leftrightarrow1-x^2\ge0\) \(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x+1\ge0\end{matrix}\right.\) \(\Leftrightarrow-1\le x\le1\)

\(\sqrt{-5x-10}\) có nghĩa \(\Leftrightarrow-5x-10\ge0\Leftrightarrow-5x\ge10\Leftrightarrow x\ge-2\)

Ta có: 

\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\)

\(=\sqrt{\left(3x^2+6x+3\right)+9}+\sqrt{\left(5x^4-10x^2+5\right)+4}\)

\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\left(1\right)\)

Ta lại có:

\(-2x^2-4x+3=-2\left(x+1\right)^2+5\le5\left(2\right)\)

Từ (1) và (2) dấu = xảy ra khi \(x=-1\)

Bài 1:

a) Để căn thức \(\sqrt{\frac{2}{9-x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\frac{2}{9-x}\ge0\\9-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9-x>0\\x\ne9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 9\\x\ne9\end{matrix}\right.\Leftrightarrow x< 9\)

b) Ta có: \(x^2+2x+1\)

\(=\left(x+1\right)^2\)

mà \(\left(x+1\right)^2\ge0\forall x\)

nên \(x^2+2x+1\ge0\forall x\)

Do đó: Căn thức \(\sqrt{x^2+2x+1}\) xác được với mọi x

c) Để căn thức \(\sqrt{x^2-4x}\) có nghĩa thì \(x^2-4x\ge0\)

\(\Leftrightarrow x\left(x-4\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-4\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x< 0\end{matrix}\right.\)

Bài 3:

a) Ta có: \(\sqrt{\left(3-\sqrt{10}\right)^2}\)

\(=\left|3-\sqrt{10}\right|\)

\(=\sqrt{10}-3\)(Vì \(3< \sqrt{10}\))

b) Ta có: \(\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=\left|\sqrt{5}-2\right|\)

\(=\sqrt{5}-2\)(Vì \(\sqrt{5}>2\))

c) Ta có: \(3x-\sqrt{x^2-2x+1}\)

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

\(=3x-\left|x-1\right|\)

\(=\left[{}\begin{matrix}3x-\left(x-1\right)\left(x\ge1\right)\\3x-\left(1-x\right)\left(x< 1\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}3x-x+1\\3x-1+x\end{matrix}\right.=\left[{}\begin{matrix}2x+1\\4x-1\end{matrix}\right.\)

a) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)

Để A nguyên thì 4 ⋮ √x - 2

\(\Rightarrow\sqrt{x}-2\inƯ\left(4\right)\)

\(\Rightarrow\sqrt{x}-2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0;6;-2\right\}\)

Mà x \(\sqrt{x}\ge0\)

=> x thuộc {9; 1; 16; 0; 36}

b) 

1:

a: ĐKXĐ: 1-x>=0

=>x<=1

b: ĐKXĐ: 2/x>=0

=>x>0

c: ĐKXĐ: 4/x+1>=0

=>x+1>0

=>x>-1

d: ĐKXĐ: x^2+2>=0

=>x thuộc R

Câu 2:

a: \(=\left|-\sqrt{2-1}\right|=\sqrt{1}=1\)

b: \(=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)

\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)

\(=32+8\sqrt{15}-8\sqrt{15}-30\)

=2