ĐK: \(\hept{\begin{cases}x^2+4x+4\ne0\\4-x^2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\ne0\\\left(2-x\right)\left(2+x\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}}\)

\(P=\frac{x^3-4x}{x^2+4}.\left(\frac{1}{x^2+4x+4}+\frac{1}{4-x^2}\right)\)

\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{1}{\left(x+2\right)^2}+\frac{-1}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{x-2-\left(x+2\right)}{\left(x+2\right)^2\left(x-2\right)}\right)\)

\(=\frac{x\left(x-2\right)\left(x+2\right)}{x^2+4}.\frac{-4}{\left(x+2\right)^2\left(x-2\right)}=\frac{-4x}{\left(x^2+4\right)\left(x+2\right)}\)

Mình ko thêm bớt hạng tử nhé.

\(8x^3-3x+6x^2-1\)

\(=\left(8x^3-1\right)+\left(6x^2-3x\right)\)

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

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

\(=\left(2x-1\right)\left(4x^2+5x+1\right)\)

\(=\left(2x-1\right)\left[4x\left(x+1\right)+\left(x+1\right)\right]\)

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

\(8x^3-3x+6x^2-1=\left(8x^3-12x^2+6x-1\right)+\left(18x^2-9x\right)\)

\(=\left(\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\right)+\left(18x^2-9x\right)\)

\(=\left(2x-1\right)^3+9x\left(2x-1\right)=\left(2x-1\right)\left(\left(2x-1\right)^2+9x\right)\)

\(=\left(2x-1\right)\left(4x^2-4x+1+9x\right)=\left(2x-1\right)\left(4x^2+5x+1\right)\)

\(x^2-xy=6x-5y-8\)

\(\Rightarrow x^2-xy-6x+5y+8=0\)

\(\Rightarrow\left(x^2-xy-x\right)-\left(5x-5y-5\right)+3=0\)

\(\Rightarrow x\left(x-y-1\right)-5\left(x-y-1\right)=-3\)

\(\Rightarrow\left(x-y-5\right)\left(x-1\right)=-3\)

Từ đó bạn tìm ước thì ra kết quả.Chúc bạn học tốt.

đặt \(x-y=k\)

\(x^2-xy=6x-5y-8\Rightarrow x\left(x-y\right)=x+\left(5x-5y\right)-8\Rightarrow xk=x+5\left(x-y\right)-8\)

\(\Rightarrow xk=x+5k-8\Rightarrow xk=x+5k-5-3\Rightarrow xk-x-5k+5=-3\)

\(\Rightarrow x\left(k-1\right)-5\left(k-1\right)=3\Rightarrow\left(x-5\right)\left(k-1\right)=3\Rightarrow x-5;k-1\inƯ\left(-3\right)=+-1;+-3\)

nếu \(x-5=1\Rightarrow x=6\)thì \(k-1=-3\Rightarrow k=-2\Rightarrow y=x-k=6-\left(-2\right)=8\)

nếu \(x-5=3\Rightarrow x=8\)thì \(k-1=-1\Rightarrow k=0\Rightarrow y=x-k=8-0=8\)

nếu \(x-5=-1\Rightarrow x=4\)thì \(k-1=3\Rightarrow k=4\Rightarrow y=x-k=4-4==0\)

nếu \(x-5=-3\Rightarrow x=2\)thì \(k-1=1\Rightarrow k=2\Rightarrow y=x-k=2-2=0\)

vậy (x;y)=(6;8) (8;8) (4;0) (2;0)

\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)

\(=\frac{\left(x+3\right).\left(x-3\right)}{x.\left(x-3\right)}-\frac{x^2}{\left(x-3\right).x}+\frac{9}{x.\left(x-3\right)}=\frac{x^2-9-x^2+9}{x.\left(x-3\right)}=0\)

Sửa đề \(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

\(\Leftrightarrow x^2-2xy+y^2+z^2-2zx+x^2+y^2-2yz+z^2=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(z-x\right)^2+\left(y-z\right)^2=0\)

Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(z-x\right)^2\ge0\\\left(y-z\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)

Dấu "=" xảy ra khi x = y = z

Với x = y =z thay vào pt ban đầu ta đc \(x^2+x^2+x^2=x.x+x.x\)

                                                 \(\Leftrightarrow3x^2=2x^2\)

                                                 \(\Leftrightarrow x^2=0\) 

                                                 \(\Leftrightarrow x=y=z=0\)

\(x^5y-xy^5=xy\left(x^4-y^4\right)\)

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

                      \(=xy\left(x^4-1\right)-xy\left(y^4-1\right)\)

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

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

Xét \(xy\left(x-1\right)\left(x+1\right)\left(x^2+1\right)=xy\left(x-1\right)\left(x+1\right)\left(x^2-4+5\right)\)

 \(=xy\left(x-1\right)\left(x+1\right)\left(x^2-4\right)+5xy\left(x-1\right)\left(x+1\right)\)

\(=y.\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)+5y\left(x-1\right)x\left(x+1\right)\)

Do x-2 ; x-1 ; x ; x+1 ; x+2 là 5 số liên tiếp

\(\Rightarrow\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)⋮2;3;5\)

Mà (2;;3;5) = 1

\(\Rightarrow\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)⋮\left(2.3.5=30\right)\)

\(\Rightarrow y\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)⋮30\)

Lại có \(5\left(x-1\right)x\left(x+1\right)⋮2;3;5\Rightarrow5\left(x-1\right)x\left(x+1\right)⋮30\)

                                                          \(\Rightarrow5y\left(x-1\right)x\left(x+1\right)⋮30\)

Do đó \(y\left(x-2\right)\left(x-1\right)x\left(x+1\right)\left(x+2\right)-5y\left(x-1\right)x\left(x+1\right)⋮30\)

\(\Rightarrow xy\left(x-1\right)\left(x+1\right)\left(x^2+1\right)⋮30\)

Tương tự \(xy\left(y-1\right)\left(y+1\right)\left(y^2+1\right)⋮30\)

\(\Rightarrow xy\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-xy\left(y-1\right)\left(y+1\right)\left(y^2+1\right)⋮30\)

\(\Rightarrow x^5y-xy^5⋮30\)

Mạnh ê,tôi vào đc nixk này rồi hehe

Duy thoát ra ngay đi

a, ĐK: \(\hept{\begin{cases}x+2\ne0\\x\ne0\end{cases}\Rightarrow}\hept{\begin{cases}x\ne-2\\x\ne0\end{cases}}\)

b, \(B=\left(1-\frac{x^2}{x+2}\right).\frac{x^2+4x+4}{x}-\frac{x^2+6x+4}{x}\)

\(=\frac{-x^2+x+2}{x+2}.\frac{\left(x+2\right)^2}{x}-\frac{x^2+6x+4}{x}\)

\(=\frac{\left(-x^2+x+2\right)\left(x+2\right)-\left(x^2+6x+4\right)}{x}\)

\(=\frac{-x^3-2x^2+x^2+2x+2x+4-\left(x^2+6x+4\right)}{x}\)

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

c, x = -3 thỏa mãn ĐKXĐ của B nên với x = -3 thì 

\(B=-\left(-3\right)^2-2.\left(-3\right)-2=-9+6-2=-5\)

d, \(B=-x^2-2x-2=-\left(x^2+2x+1\right)-1=-\left(x+1\right)^2-1\le-1\forall x\)

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của B là - 1 khi x = -1

Thanks bạn ;)