Ta có: \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b-c\right)\left(b^2+bc+c^2\right)+bc^3-a^3b+a^3c-b^3c\)

\(=a\left(b-c\right)\left(b^2+bc+c^2\right)-bc\left(b-c\right)\left(b+c\right)-a^3\left(b-c\right)\)

\(=\left(b-c\right)\left(ab^2+abc+c^2a-b^2c-bc^2-a^3\right)\)

\(=\left(b-c\right)\left[c^2\left(a-b\right)-a\left(a-b\right)\left(a+b\right)+bc\left(a-b\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(c^2-a^2-ab+bc\right)\)

\(=\left(a-b\right)\left(b-c\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)

Đặt \(2020-x=u;x-2021=v\)thì \(u+v=-1\)

Phương trình trở thành \(\frac{u^2+uv+v^2}{u^2-uv+v^2}=\frac{19}{49}\Leftrightarrow30u^2+30v^2+68uv=0\)

\(\Leftrightarrow15\left(u+v\right)^2+4uv=0\Leftrightarrow4uv=-15\Leftrightarrow uv=\frac{-15}{4}\)

hay \(\left(2020-x\right)\left(x-2021\right)=-\frac{15}{4}\Leftrightarrow x^2-4041x+4082416,25=0\)

Dùng công thức nghiệm tìm được x = 2022, 5 hoặc x = 2018, 5

Câu 1:

\(a^3+a^2b-ab^2-b^3\)

\(=a^2\left(a+b\right)-b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-b^2\right)\)

\(=\left(a+b\right)\left(a-b\right)\left(a+b\right)\)

\(=\left(a+b\right)^2\left(a-b\right)\)

Câu 2:

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+bc^3-a^3b+a^3c-b^3c\)

\(=a\left(b-c\right)\left(b^2+bc+c^2\right)-a^3\left(b-c\right)-bc\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(ab^2+abc+c^2a-a^3-b^2c-bc^2\right)\)

\(=\left(b-c\right)\left[a\left(c-a\right)\left(c+a\right)-b^2\left(c-a\right)-bc\left(c-a\right)\right]\)

\(=\left(b-c\right)\left(c-a\right)\left(ca+a^2-b^2-bc\right)\)

\(=\left(b-c\right)\left(c-a\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)

\(ĐK:-3\le x\le6\)

Đặt \(t=\sqrt{x+3}+\sqrt{6-x}\left(t>0\right)\Rightarrow t^2=9+2\sqrt{\left(x+3\right)\left(6-x\right)}\Rightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=\frac{t^2-9}{2}\)

Phương trình trở thành \(t-\frac{t^2-9}{2}=3\Leftrightarrow t^2-2t-3=0\Leftrightarrow\left(t-3\right)\left(t+1\right)=0\Leftrightarrow\orbr{\begin{cases}t=3\left(tm\right)\\t=-1\left(L\right)\end{cases}}\)    

Với t = 3 thì \(\sqrt{x+3}+\sqrt{6-x}=3\Rightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\Rightarrow\orbr{\begin{cases}x=6\\x=-3\end{cases}}\left(tm\right)\)  

Vậy phương trình có tập nghiệm S = {6; -3}

Câu 1:

\(a^2+2ab+b^2-ac-bc\)

\(=\left(a+b\right)^2-c\left(a+b\right)\)

\(=\left(a+b\right)\left(a+b-c\right)\)

Câu 2:

\(5x^2-5y^2-10x+10y\)

\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+5y-10\right)\)

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

Câu 3:

\(3x^2-6xy+3y^2-12z^2\)

\(=3\left[\left(x-y\right)^2-4z^2\right]\)

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

Câu 4:

\(x^4+x^3+x^2-1\)

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

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

Câu 5:

\(x^3-3x^2+3x-1-y^3\)

\(=\left(x-1\right)^3-y^3\)

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

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

Câu 6:

\(x^4-x^2+2x-1\)

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

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

Câu 7:

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

\(=\left(x+y\right)^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\)

\(=3xy\left(x+y\right)\)