x^n-2( x³ + x² - 5 )

= xⁿ⁺¹ + xⁿ - 5x^n-2

c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

a) \(=15x^3-6x^2+3x\)

b) \(=x^3-8\)

c) \(=x^2+10x+25\)

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

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

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

\(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)=x^3-5x^2+x-2x^2+10x-2-x^3-11x=-7x^2-2\)

\(a,=27x^3+27x^2+9x+1\)

\(b,=\dfrac{x^3}{27}-\dfrac{x^2}{3}+x-1\)

\(c,=-\left(27x^3-27x^2y^2+9xy^4-y^6\right)\)

\(=-27x^3+27x^2y^2-9xy^4+y^6\)

\(d,=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)

a) \(\left(3x+1\right)^3=27x^3+27x^2+9x+1\)

b) \(\left(\dfrac{x}{3}-1\right)^3=\dfrac{x^3}{27}-\dfrac{x^2}{3}\)

c) \(\left(-y^2+3x\right)^3=27x^3-27x^2y^2+9xy^4-y^6\)

d) \(\left(\dfrac{x}{y}-\dfrac{2y}{x}\right)^3=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)

\(\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\)

\(TH1:x-1=0\Leftrightarrow x=1\)

\(TH2:x+1=0\Leftrightarrow x=-1\)

\(TH3:x+2=0\Leftrightarrow x=-2\)

nhân đa thức vs đa thức , ko phải tìm x đâu bạn ạ! dù sao cững cảm ơn nh!