a, đk \(\left\{{}\begin{matrix}x^2-4x+4\ge0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\left(dung\right)\\x\ne3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\x\ne3\end{matrix}\right.\)

b, đk \(\left\{{}\begin{matrix}2x-3>0\\x-7>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>7\end{matrix}\right.\Leftrightarrow x>7\)

c, đk \(x-10\sqrt{x}+25>0\Leftrightarrow\left(\sqrt{x}-5\right)^2>0\Rightarrow x\ne25\)

\(7-4\sqrt{3}=7-2.2\sqrt{3}=4-2.2\sqrt{3}+3=\left(2-\sqrt{3}\right)^2=\left(\sqrt{3}-2\right)^2\)

a, \(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

b, \(=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)

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

ĐKXĐ : 

\(\left\{{}\begin{matrix}\sqrt{a}+2\ne0\\\sqrt{a}-2\ne0\\\sqrt{a}\ne0\\\sqrt{a}x\text{đ}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{a}\ne2\\a\ne0\\a\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\ne4\\a>0\end{matrix}\right.\)

Rút gọn :

\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)

\(=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\dfrac{a-4}{\sqrt{a}}\)

\(=\dfrac{\left(\sqrt{a}-2+\sqrt{a}+2\right)\left(\sqrt{a}-2-\sqrt{a}-2\right)}{\left(\sqrt{a}\right)^2-2^2}.\dfrac{a-4}{\sqrt{a}}\)

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

\(=2.\left(-4\right)=-8\)

\(x,y>0;n\in N,n>2\).

- Áp dụng bất đẳng thức Caushy ta có:

\(x^n+x^n+...+x^n+y^n\) (có \(\left(n-1\right)\) \(x^n\)) \(\ge n\sqrt[n]{x^{n\left(n-1\right)}.y^n}=n.x^{n-1}.y\)

\(\Rightarrow\left(n-1\right)x^n+y^n\ge n.x^{n-1}.y^n\left(1\right)\) 

- Tương tự: \(\left(n-1\right)y^n+x^n\ge n.y^{n-1}.x\left(2\right)\)

- Từ (1), (2) suy ra:

\(n\left(x^n+y^n\right)\ge n.xy\left(x^{n-2}+y^{n-2}\right)\)

\(\Rightarrow\left(x^n+y^n\right)\ge xy\left(x^{n-2}+y^{n-2}\right)\left(đpcm\right)\)

- Dấu "=" xảy ra \(\Leftrightarrow x^n=y^n;x,y>0\Leftrightarrow x=y\)

Cách khác:
$x^n+y^n\geq xy(x^{n-2}+y^{n-2})$

$\Leftrightarrow x^n+y^n-x^{n-1}y-xy^{n-1}\geq 0$

$\Leftrightarrow (x^n-x^{n-1}y)+(y^n-xy^{n-1})\geq 0$

$\Leftrightarrow x^{n-1}(x-y)-y^{n-1}(x-y)\geq 0$

$\Leftrightarrow (x-y)(x^{n-1}-y^{n-1})\geq 0$

$\Leftrightarrow (x-y)(x-y)(x^{n-2}+x^{n-3}y+...+y^{n-2})\geq 0$

$\Leftrightarrow (x-y)^2(x^{n-2}+x^{n-3}y+...+y^{n-2})\geq 0$

(luôn đúng với mọi $x,y>0$ và $n\in\mathbb{N}>2$)

Ta có đpcm

Dấu "=" xảy ra khi $x=y$

\(=\dfrac{4+a+4\sqrt{a}-a+6\sqrt{a}-9}{2a-\sqrt{a}}=\)

\(=\dfrac{10\sqrt{a}-5}{2a-\sqrt{a}}=\dfrac{5\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(2\sqrt{a-1}\right)}=\)

\(=\dfrac{5\sqrt{a}}{a}\)

Lời giải:

ĐKXĐ: $a>0; a\neq \frac{1}{4}$
\(A=\frac{(2+\sqrt{a}-\sqrt{a}+3)(2+\sqrt{a}+\sqrt{a}-3)}{\sqrt{a}(2\sqrt{a}-1)}=\frac{5(2\sqrt{a}-1)}{\sqrt{a}(2\sqrt{a}-1)}=\frac{5}{\sqrt{a}}\)

P/s: Lần sau bạn lưu ý ghi đầy đủ yêu cầu đề bài