Ta có : 

\(\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)

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

\(-\frac{\sqrt{a}+3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

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

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

\(=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)

Ta có : 

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)

\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)

Ta có : 

\(18-\sqrt{128}=18-8\sqrt{2}=16-2.4.\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)

Vậy 

\(\sqrt{18-\sqrt{128}}=4-\sqrt{2}\)

Thay vào ta có

\(\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)

\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)

\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)

Lại có : 

\(4+2\sqrt{3}=3+2.1.\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)

Do đó : 

\(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Vậy : 

\(\sqrt{6-2\sqrt{4+2\sqrt{3}}}=\sqrt{6-2\left(\sqrt{3}+1\right)}\)

\(=\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{3-2.1.\sqrt{3}+1}\)

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

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

Vậy : \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{3}-1\)

Đặt \(\sqrt{2-x}=a\)

Ta có:

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

\(=\left(3-t^2\right)t=3t-t^3\)

Đến đây đạo hàm thử xem ?

\(5+2\sqrt{3}=5+2\sqrt{3}\)

cậu thử tính máy tính

!!!~~~~~~~

à quên, đề là phân tích đa thức thành nhân tử

a) Ta có bảng giá trị tương ứng x và y sau :

Quả parabol hơi xấu tí thông cảm =))

Mục tiêu -500 sp mong giúp đỡ

ta có

\(0\le\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\left(\forall x,y,z>0\right)\)

\(\Leftrightarrow2xy+2yz+2zx\le2\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\)(1)

dấu  = xảy ra khi

\(x=y=z=0\)

theo giả thiết ta có

\(x\left(x+1\right)+y\left(y+1\right)+z\left(z+1\right)\le18\)

\(\Leftrightarrow x^2+y^2+z^2\le18-\left(x+y+z\right)\left(2\right)\)

từ (1) zà (2) suy ra

\(\left(x+y+z\right)^2\le54-3\left(x+y+z\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-54\le0\)

\(\Leftrightarrow\left(x+y+z-6\right)\left(x+y+z+9\right)\le0\)

\(\Leftrightarrow0< x+y+z\le6\left(do\left(x+y+z>0;9>0\right)\right)\)

áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có

\(P=\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\ge\frac{9}{2\left(x+y+z\right)+3}\ge\frac{9}{2.6+3}=\frac{3}{5}\)

Dấu = xảy ra khi zà chỉ khi

\(\hept{\begin{cases}x+y+1=y+z+1=z+x+1\\x+y+z=6\end{cases}=>x=y=z=2}\)

zậy MinP= 3/5 khi x=y=z=2

Ta có : x(x + 1) + y (y+1 ) + z(z + 1) \(\le18\)

<=> x² + y² + z² + ( x + y + z ) \(\le18\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

=> 54 \(\ge\)( x + y+z)² + 3(x + y + z) 

<=> -9 \(\le\)x + y + z \(\le\)6

=> 0 \(\le\)x+y+z \(\le\)6 

\(\frac{1}{x+y+1}+\frac{x+y+1}{25}\ge\frac{2}{5}\)

\(\frac{1}{y+z+1}+\frac{y+z+1}{25}\ge\frac{2}{5}\)

\(\frac{1}{z+x+1}+\frac{z+x+1}{25}\ge\frac{2}{5}\)

=> \(P+\frac{2\left(x+y+z\right)+3}{25}\ge\frac{6}{5}\)

=> P \(\ge\frac{27}{25}-\frac{2}{25}\left(x+y+z\right)\ge\frac{15}{25}=\frac{3}{5}\)

Dấu " =" xảy ra khi :

\(\hept{\begin{cases}x=y=z>0;x+y+z=6\\\left(x+y+1\right)^2=\left(y+z+1\right)^2=\left(z+x+1\right)^2=25\end{cases}\Leftrightarrow x=y=z=2}\)

Vậy GTNN của P là \(\frac{3}{5}\)khi x = y =z =2