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

\(\dfrac{2}{\sqrt{5}-1}-\dfrac{2}{\sqrt{5}+1}\)

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

\(=\dfrac{2.2}{\left(\sqrt{5}\right)^2-1^2}\)

\(=\dfrac{4}{4}\)

`=1`

\(\dfrac{2}{\sqrt{5}-1}-\dfrac{2}{\sqrt{5}+1}=\dfrac{2\left(\sqrt{5}+1\right)}{4}-\dfrac{2\left(\sqrt{5}-1\right)}{4}=\dfrac{2\sqrt{5}+2-2\sqrt{5}+2}{4}=1\)

\(2\sqrt{144}+\sqrt{100}-\sqrt{81}\)

\(=2\sqrt{12^2}+\sqrt{10^2}-\sqrt{9^2}\)

\(=2.12+10-9\)

\(=25\)

a) \(\left(10+9\right)x-\left(5x-1\right)\left(2x+3\right)+5x-7\)

\(=19x-\left(10x^2+15x-2x-3\right)+5x-7\)

\(=19x-10x^2-13x+3+5x-7\)

\(=-10x^2+11x-4\)

b) \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x-1\right)\left(x+1\right)\)

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

\(=4x-3x^2+3\)

c) \(\left(2x+1\right)^2-4\left(x-2\right)\left(x+2\right)-\left(x-3\right)^2\)

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

\(=\left(3x-2\right)\left(x+4\right)-4x^2+16\)

\(=3x^2+12x-2x-8-4x^2+16\)

\(=-x^2+10x+8\)

\(2^{2n+1}+5^{2n}=2^{2n}.2+25^n=4^n.2+25^n\)

- Vì số chính phương khi chia cho 3 thì chỉ có thể dư \(0\) hoặc \(1\).

\(\Rightarrow\left\{{}\begin{matrix}4^nmod3\in\left\{0;1\right\}\\25^nmod3\in\left\{0;1\right\}\end{matrix}\right.\).

Mà \(4^n,25^n\) không chia hết cho \(3\) với mọi \(n\in N\)

\(\Rightarrow\left\{{}\begin{matrix}4^nmod3=1\\25^nmod3=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(4^n.2\right)mod3=2\\25^nmod3=1\end{matrix}\right.\Rightarrow\left(4^n.2+25^n\right)⋮3\)

\(\Rightarrowđpcm\)

Lấy \(H,I,K\) trên \(AB,BC,CA\) sao cho \(DH,EI,FK\) tương ứng vuông góc với \(AB,BC,CA\). Lấy \(G\) đối xứng với \(E\) qua \(BC\). Gọi \(AG\) cắt \(DF\) tại \(J\).

Ta thấy \(\Delta BGC=\Delta BEC\sim\Delta BDA\sim\Delta AFC\). Suy ra \(\Delta BDG~\Delta BAC\sim GFC\). Từ đây \(\dfrac{DG}{AC}=\dfrac{BD}{BA}=\dfrac{AF}{AC}\Rightarrow DG=AF\). Tương tự thì \(FG=AD\). Do đó \(ADGF\) là hình bình hành. Suy ra \(J\) là trung điểm của \(DF,AG.\) 

Ta có \(IK||AB\) do \(\dfrac{IB}{IC}=\dfrac{EB^2}{EC^2}=\dfrac{AF^2}{AC^2}=\dfrac{KA}{KC}\) và \(IH||AC\)

Suy ra \(\Delta DHI\sim\Delta IKF\) vì \(\widehat{DHI}=\widehat{IKF}=90^0+\widehat{BAC}\) và \(\dfrac{DH}{IK}=\dfrac{DH}{AH}=\dfrac{AK}{KF}=\dfrac{HI}{KF}\) . Do vậy:\(\widehat{DIF}=\widehat{DIH}+\widehat{HIK}+\widehat{KIF}=\widehat{DIH}+\widehat{BAC}+\widehat{DHI}=180^0-\left(90^0+\widehat{BAC}\right)+\widehat{BAC}=90^0\)Xét \(\Delta AGE\) có đường trung bình \(IJ\). Suy ra \(AE=2IJ\)

Xét \(\Delta DIF\) có \(\widehat{DIF}=90^0\), \(J\) là trung điểm của \(DF\). Suy ra \(DF=2IJ\)

Vậy \(DF=AE.\)