\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

Ta có: \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)

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

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

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

\(\left(x+2\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

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

\(=2x^2-4x\)

\(=x^2-4x+4+3x-x^2+3-x=-2x+7\)

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

Ta có: \(\left(x-2\right)^2+\left(3-x\right)\left(x-1\right)\)

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

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

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

=4-3

=1

\(\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)=x^2+4x+4-\left(x^2+x-3x-3\right)=x^2+4x+4-x^2-x+3x+3=6x+7.\)

\(\left(5x^4-x^3+3x^2\right)\)

Tưởng đy học bài ?

Ta có: \(\left(x^3+4x^2+x-2\right):\left(x+1\right)\)

\(=\frac{x^3+x^2+3x^2+3x-2x-2}{x+1}\)

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

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

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

\(=x^3-y^3\)