a) ĐKXĐ : \(\hept{\begin{cases}x-3\ne0\\x^2-9\ne0\\x+3\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne3\\x\ne\pm3\\x\ne-3\end{cases}}\Rightarrow x\ne\pm3\)

b) A = \(\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}=\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)

Khi x = 3 => Không thỏa mãn ĐKXĐ

=> Không tồn tại A khi x = 3

a, Điều kiện xác định là : 

\(\hept{\begin{cases}x-3\ne0\\x^2-9\ne0\\x+3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne3\\\left(x-3\right)\left(x+3\right)\ne\\x\ne-3\end{cases}}0\Rightarrow x\ne\pm3}\)

Vậy \(x\ne\pm3\)

b, \(A=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)

\(=\frac{3}{x-3}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\)

\(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)

Thay x = 3 ( ktm đkxđ )

Ko tồn tại x 

Ta có đánh giá quen thuộc: \(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=3\left(x+y+z\right)\)(Do xyz = 1)\(\Rightarrow\frac{1}{x+y+z}\ge\frac{3}{\left(xy+yz+zx\right)^2}\)

Như vậy, ta cần chứng minh: \(\frac{3}{\left(xy+yz+zx\right)^2}+\frac{1}{3}\ge\frac{2}{xy+yz+zx}\)

Đặt \(t=\frac{1}{xy+yz+zx}\)thì bất đẳng thức trở thành \(3t^2+\frac{1}{3}\ge2t\Leftrightarrow9t^2+1\ge6t\Leftrightarrow\left(3t-1\right)^2\ge0\)*đúng*

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi \(\hept{\begin{cases}t=\frac{1}{xy+yz+zx}=\frac{1}{3}\\x=y=z>0,xyz=1\end{cases}}\Leftrightarrow x=y=z=1\)

P = \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}-3=y.\frac{y}{x+y}+z.\frac{z}{y+z}+x.\frac{x}{z+x}-3\)

\(=y.\left(\frac{y}{x+y}-1+1\right)+z\left(\frac{z}{y+z}-1+1\right)+x\left(\frac{x}{z+x}-1+1\right)-3\)

\(=y\left(\frac{-x}{x+y}+1\right)+z\left(\frac{-y}{y+z}+1\right)+x\left(\frac{-z}{x+z}+1\right)-3\)

\(=x+y+z-\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{xz}{z+x}\right)-3\)

Lại có \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2017\)

\(\Rightarrow x.\frac{x}{x+y}+y.\frac{y}{y+z}+z.\frac{z}{z+x}=2017\)

=> \(x\left(\frac{x}{x+y}-1+1\right)+y\left(\frac{y}{y+z}-1+1\right)+z\left(\frac{z}{z+x}-1+1\right)=2017\)

=> \(x\left(\frac{-y}{x+y}+1\right)+y\left(\frac{-z}{y+z}+1\right)+z\left(\frac{-x}{x+z}+1\right)=2017\)

=> \(x+y+z-\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x}\right)=2017\)

Khi đó P = 2017 - 3 = 2014

a)T=\(n^2+7n+22\equiv n^2-2n+4\equiv\left(n-1\right)^2+3\left(mod9\right)\)

TH1 (n-1)² khong chia het cho 3 

=> T khong chia het cho 3 => T khong chia het cho 9 =>dpcm

TH2 (n-1)²  chia het cho 3=>  (n-1)²  chia het cho 9

=> T khong chia het cho 9 => dpcm

b)T=\(n^2-5n-49\equiv n^2+8n+3\equiv\left(n+4\right)^2-13\left(mod13\right)\)

lam tuong tu phan a)

\(2n^3+n^2+7n+1⋮2n-1\)

Suy ra : \(2n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Thử lại xem ngon chưa rồi chọn 

\(\frac{2x-5}{5x-10}=0\)ĐKXĐ : \(x\ne2\)

\(\Leftrightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\)( tm )

Vậy \(x=\frac{5}{2}\)