a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)

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

\(-2y^3\left(4x^3-xy^2+y^3\right)\)

\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)

\(-8x^3y^3+2xy^5-2y^6\)

\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)

\(-\left(x^3y^3+8x^3y^3\right)\)

\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)

b) 

(!)  \(2\left(x+y\right)^2-7\left(x+y\right)+5\)

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

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

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

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

\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)

\(=2\left(xy+yz+zx\right)\)

\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)

=> \(\left(2x+3-2x+5\right)^2=x^2+6x+64\)

=> \(8^2=x^2+6x+64\)

=> \(64=x^2+6x+64\)

=> \(x\left(x+6\right)=0\)

=> \(\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)

\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2=x^2+6x+64\)

\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)

\(\Leftrightarrow8^2=x^2+6x+64\)\(\Leftrightarrow64=x^2+6x+64\)

\(\Leftrightarrow x^2+6x+64-64=0\)\(\Leftrightarrow x^2+6x=0\)

\(\Leftrightarrow x\left(x+6\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

Vậy tập nghiệm của phương trình là: \(S=\left\{-6;0\right\}\)

b, Ta có : \(\left(x^2-x\right)^2-2=x^2-x\)

             \(\Leftrightarrow t^2-2=t\)

             \(\Leftrightarrow\orbr{\begin{cases}t=2\\t=-1\end{cases}}\)

             \(\Leftrightarrow\orbr{\begin{cases}x^2-x=2\\x^2-x=-1\end{cases}}\)

             \(\Leftrightarrow\hept{\begin{cases}x=2\\x=-1\\x\notinℝ\end{cases}}\)

             \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy \(x_1=-1;x_2=2\)

c, Ta có : \(x.\left(x+1\right).\left(x-1\right).\left(x+2\right)=24\)

             \(\Leftrightarrow x.\left(x^2-1\right).\left(x+2\right)=24\)

              \(\Leftrightarrow\left(x^3-x\right).\left(x+2\right)=24\)

              \(\Leftrightarrow x^4+2.x^3-x^2-2.x=24\)

              \(\Leftrightarrow x^4+2.x^3-x^2-2.x-24=0\)

              \(\Leftrightarrow x^4-2.x^3+4.x^3-8.x^2+7.x^2-14.x+12.x-24=0\)

              \(\Leftrightarrow x^3.\left(x-2\right)+4.x^2.\left(x-2\right)+7.x.\left(x-2\right)+12.\left(x-2\right)=0\)

              \(\Leftrightarrow\left(x-2\right).\left(x^3+4.x^2+x^2+7.x+12\right)=0\)

              \(\Leftrightarrow\left(x-2\right).\left(x^3+3.x^2+x^2+3.x+4.x+12\right)=0\)

             \(\Leftrightarrow\left(x-2\right).\left[x^2.\left(x+3\right)+x.\left(x+3\right)+4.\left(x+3\right)\right]=0\)

             \(\Leftrightarrow\left(x-2\right).\left(x+3\right).\left(x^2+x+4\right)=0\)

              \(\Rightarrow\hept{\begin{cases}x-2=0\\x+3=0\\x^2+x+4=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\\x\notinℝ\end{cases}}\)

               \(\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy \(x_1=-3;x_2=2\)

Cô ơi em có cách khác ạ :)

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)

Dấu "=" xảy ra tại x=y=z=0

Khi đó T=0

Ta có: 

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

<=> \(\left(a^2+b^2+c^2\right)\)\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

<=> \(x^2+y^2+z^2=\left(a^2+b^2+c^2\right)\frac{x^2}{a^2}+\left(a^2+b^2+c^2\right)\frac{y^2}{b^2}+\left(a^2+b^2+c^2\right)\frac{z^2}{c^2}\)

<=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)

vì a, b , c khác 0 nên \(\frac{\left(b^2+c^2\right)}{a^2};\frac{\left(c^2+a^2\right)}{b^2};\frac{\left(b^2+a^2\right)}{c^2}\ne0\)

\(\frac{\left(b^2+c^2\right)}{a^2}x^2\ge0;\frac{\left(a^2+c^2\right)}{b^2}y^2\ge0;\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x, y, z

=> \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2\ge0\)với mọi x; y; z

Do đó: \(\frac{\left(b^2+c^2\right)}{a^2}x^2+\frac{\left(a^2+c^2\right)}{b^2}y^2+\frac{\left(a^2+b^2\right)}{c^2}z^2=0\)

=> x = y = z = 0

Vậy T = 0 

Phân thức đại số: \(\frac{1}{x-3}\)