\(a,Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}};x>0;x\ne1;x\ne4\)

\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

\(=\frac{2}{x-1}\)

\(a,\)\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

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

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

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

\(b,Q=\frac{2}{x-1}\)

\(Q\in Z\Leftrightarrow\frac{2}{x-1}\in Z\Rightarrow x-1\inƯ_2\)

Mà \(Ư_2=\left\{\pm1;\pm2\right\}\)

TH1 : \(x-1=-1\Rightarrow x=0\)

TH2 : \(x-1=1\Rightarrow x=2\)

TH3 : \(x-1=-2\Rightarrow x=-1\)

TH4 :\(x-1=2\Rightarrow x=3\)

\(\Rightarrow\)x nguyên lớn nhất là 3 để Q là số nguyên

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

Ta thấy A luôn dương 

\(\Rightarrow A_{min}\Leftrightarrow\sqrt{\left(x-2\right)^2+3}\)Nhỏ nhất\(\Rightarrow\left(x-2\right)^2\)nhỏ nhất 

Hay \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

\(\Rightarrow A_{min}=\sqrt{0+3}=\sqrt{3}\Leftrightarrow x=2\)

\(B=\sqrt{x-2\sqrt{x}-3}=\sqrt{x+\sqrt{x}-3\sqrt{x}-3}\)

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

\(B_{min}\Leftrightarrow B=0\Rightarrow\sqrt{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\\sqrt{x}-3=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1\\\sqrt{x}=3\end{cases}\Rightarrow}\orbr{\begin{cases}x\in\varnothing\\x=9\end{cases}}}\)

Vậy \(B_{min}=0\Leftrightarrow x=9\)

a) \(\sqrt{21+12\sqrt{3}}=\sqrt{18+2.6.\sqrt{3}+3}\)

                                   \(=\sqrt{\left(18+\sqrt{3}\right)^2}\)

                                   \(=18+\sqrt{3}\)

b) \(\sqrt{57-40\sqrt{2}}=\sqrt{25-2.5.4\sqrt{2}+16.2}\)

                                   \(=\sqrt{\left(5-4\sqrt{2}\right)^2}\)

                                   \(=5-4\sqrt{2}\)

c) \(\sqrt{11-6\sqrt{2}}=\sqrt{9-2.3.\sqrt{2}+2}\)

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

                                 \(=3-\sqrt{2}\)

\(\sqrt{19-6\sqrt{2}}\)

\(=\sqrt{18-2.3\sqrt{2}+1}\)

\(=\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}+1}\)

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

\(=3\sqrt{2}-1\)

ĐK: \(-x^2+x+1\ge0\) (xấu quá em hok dám giải đâu:v)

PT \(\Leftrightarrow4x^2-4x+3\left(1-\sqrt{x-x^2+1}\right)=0\)

\(\Leftrightarrow4x\left(x-1\right)+3.\frac{x\left(x-1\right)}{1+\sqrt{x-x^2+1}}=0\)

\(\Leftrightarrow x\left(x-1\right)\left(4+\frac{3}{1+\sqrt{x-x^2+1}}\right)=0\)

Cái ngoặc to hiển nhiên vô nghiệm.

Do đó x = 0 (TM) hoặc x = 1 (TM)

Vậy....

P.s: đúng ko ta mà sao em thấy đơn giản quá, thường liên hợp kiểu này cái ngoặc to xấu xí lắm mà sao lần này nó dễ..

bạn làm đúng rồi nha

Ôi zời:(

Áp dụng BĐT Cauchy-Schwarz dạng Engel \(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)

Mặt khác ta có đẳng thức: \(a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\) (khai triển cái vế phải ra sẽ thấy nó bằng nhau).

Do đó \(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)

\(=\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2}{a+b+c}\)(đpcm)

Đúng ko ta?:3

bài này max ping phải là \(\ge\frac{a+b+c}{3}\) chứ nhỉ :) 

\(\frac{a^3}{a^2+ab+b^2}=\frac{2a^3}{3\left(a^2+b^2\right)-\left(a-b\right)^2}\ge\frac{2a^3}{3\left(a^2+b^2\right)}\)

\(=\frac{2}{3}\left(a-\frac{ab^2}{a^2+b^2}\right)\ge\frac{2}{3}\left(a-\frac{ab^2}{2ab}\right)=\frac{2}{3}\left(a-\frac{b}{2}\right)\)

tương tự cộng lại ta có: \(\frac{2}{3}\left(a+b+c-\frac{a}{2}-\frac{b}{2}-\frac{c}{2}\right)=\frac{a+b+c}{3}\)

Dấu "=" xảy ra khi a=b=c 

Ta có 

\(\left(a^2+2\right)\left(b^2+2\right)=\left(a^2+1\right)\left(b^2+1\right)+a^2+b^2+3\ge\left(a+b\right)^2+\frac{\left(a+b\right)^2}{2}+3=\frac{3}{2}\left[\left(a+b\right)^2+2\right]\)

\(\Rightarrow VT\ge\frac{3}{2}\left[\left(a+b\right)^2.c^2+4+2\left(a+b\right)^2+2c^2\right]\)

           \(\ge\frac{3}{2}\left[4\left(a+b\right)c+2\left(a+b\right)^2+2c^2\right]=VP\)

=> ĐPCM

Dấu "=" xảy ra khi 

\(a=b=c=\frac{\pm1}{\sqrt{2}}\)

Đề PBC 2015-2016 nè

\(0< a,b,c< 1\)\(\Rightarrow\)\(\hept{\begin{cases}ab< a;a^2< a\\bc< b;b^2< b\\ca< c;c^2< c\end{cases}}\)

\(a\ge b\ge c\)

\(\frac{1}{3}\le a< 1\Rightarrow\left(a-\frac{1}{3}\right)\left(a-1\right)\le0\)

\(\Rightarrow\)\(\frac{3}{ab+bc+ca}+\frac{2}{a^2+b^2+c^2}\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(a+b+c\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)

aaaaaaaa bỏ mấy đoạn trên đi nha >_< vẽ bùa đó, lấy mỗi đoạn dưới thôi