\(a,b,c\ge1\) chứ nhỉ?

\(a^5+b^5+c^5\ge a+b+c\)

\(\Leftrightarrow a^5+b^5+c^5-a-b-c\ge0\)

\(\Leftrightarrow a\left(a^4-1\right)+b\left(b^4-1\right)+c\left(c^4-1\right)\ge0\)

- Điều này đúng do \(a,b,c\ge1\)

- Vậy BĐT đã được c/m.

Với \(a > 0,a \ne 1\) có:

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

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

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

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

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

Ta có \(\sqrt{16-6\sqrt{7}}=\sqrt{9-2.3.\sqrt{7}+7}\) \(=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7}\)

Và \(\sqrt{29-4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}+1}\) \(=\sqrt{\left(2\sqrt{7}-1\right)^2}=2\sqrt{7}-1\)

Do đó biểu thức đã cho bằng \(\left(3-\sqrt{7}\right)-\left(2\sqrt{7}-1\right)=4-3\sqrt{7}\)

\(\dfrac{1}{P}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{2.\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=2-\dfrac{2}{\sqrt{x}+1}\)(\(x>0;\sqrt{x}+1>1\))

\(\dfrac{1}{P}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\) (do mẫu thức lớn hơn 1 nên có thế làm  theo cách này)theo điều kiện ta chỉ có 1 TH:

\(\sqrt{x}+1=2\Leftrightarrow x=1\left(TM\right)\)

vậy.............

\(P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

\(\Rightarrow\dfrac{1}{P}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{2\sqrt{x}+2-2}{\sqrt{x}+1}=2-\dfrac{2}{\sqrt{x}+1}\)

Để \(\dfrac{1}{P}\) nguyên

\(\Leftrightarrow\sqrt{x}+1\in\text{Ư}\left(2\right)\)

Ta có bảng : 

bài này mình nghĩ bạn nên đặt  \(\sqrt{x}=a\) và \(\sqrt{x-1}=b\)

Xét tg vuông ADN và tg vuông DCM có

AD=CD (cạnh hình vuông) (1)

Ta có

CD=BC (cạnh hình vuông)

NC=ND; MB=MC (gt)

=> ND=MC=MB=BC/2 (2)

Từ (1) và (2) => tg ADN = tg DCM (Hai tg vuông có 2 cạnh góc vuông bằng nhau) \(\Rightarrow\widehat{DAN}=\widehat{CDM}\)

Mà \(\widehat{CDM}+\widehat{ADM}=\widehat{ADC}=90^o\)

\(\Rightarrow\widehat{DAN}+\widehat{ADM}=90^o\)

Xét tg ADH có

\(\widehat{DAN}+\widehat{ADM}=90^o\Rightarrow\widehat{AHD}=90^o\Rightarrow AN\perp DM\)

b/

Xét tg vuông ADN có

\(DN=\dfrac{CD}{2}=\dfrac{AB}{2}=\dfrac{2}{2}=1\)

\(AN=\sqrt{AD^2+DN^2}=\sqrt{2^2+1^2}=\sqrt{5}\) (Pitago)

\(DN^2=NH.AN\) (trong tg vuông bình phương 1 cạnh góc vuông bằng tích giữa hình chiếu cạnh góc vuông đó trên cạnh huyền với cạnh huyền)

\(\Rightarrow NH=\dfrac{DN^2}{AN}=\dfrac{1^2}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)

\(\Rightarrow AH=AN-NH=\sqrt{5}-\dfrac{\sqrt{5}}{5}=\dfrac{4\sqrt{5}}{5}\)

Xét tg vuông ADN và tg vuông ABM có

AD=AB (cạnh hình vuông)

ND=MB (cmt)

=> tg ADN = tg ABM (Hai tg vuông có 2 cạnh góc vuông bằng nhau)

\(\Rightarrow\widehat{DAN}=\widehat{BAM}=\alpha\)

Ta có \(\widehat{MAN}=\widehat{BAD}-\widehat{DAN}-\widehat{BAM}=\dfrac{\Pi}{2}-2\alpha\)

\(\Rightarrow\cos\widehat{MAN}=\cos\left(\dfrac{\Pi}{2}-2\alpha\right)=\sin2\alpha=2\sin\alpha.\cos\alpha\)

Mà 

\(\sin\alpha=\dfrac{DN}{AN}=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5};\cos\alpha=\dfrac{AD}{AN}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

\(\Rightarrow\cos\widehat{MAN}=2.\dfrac{\sqrt{5}}{5}.\dfrac{2\sqrt{5}}{5}=\dfrac{4}{5}=0,8\)

\(P=\dfrac{13}{3}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{13}{3}=0\left(\text{Đ}KC\text{Đ}:x>0\right)\)

\(\Leftrightarrow\dfrac{3x+3\sqrt{x}+3-13\sqrt{x}}{3\sqrt{x}}=0\)

\(\Rightarrow3x-10\sqrt{x}+3=0\)

Đặt \(\sqrt{x}=t\left(t>0\right)\)

\(\Rightarrow3t^2-10t+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{1}{3}\end{matrix}\right.\)( TM)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{9}\end{matrix}\right.\)(TM)

\(ĐK:x>0\\ P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\dfrac{13}{3}< =>3x+3\sqrt{x}+3=13\sqrt{x}\\ < =>3x-10\sqrt{x}+3=0\\ < =>\left(\sqrt{x}-3\right)\left(3\sqrt{x}-1\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\3\sqrt{x}-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=9\\x=\dfrac{1}{9}\end{matrix}\right.\left(TMDK\right)}}\)

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

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

\(=\dfrac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\sqrt{2}+1\)

\(\sqrt{2}T=\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{5}+1}-2+\sqrt{2}\)

\(=\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}+1}-2+\sqrt{2}\)

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

\(T=\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

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

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

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

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

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

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