a) Đặt 3x làm nhân tử chung ta được 3x( 8 x 4   +   3 x 2  + 6x).

Thực hiện phép chia được thương  8 x 4   -   3 x 2  + 6x.

b) Thực hiện phép chia từng đơn thức được kết quả  5 3 x 2 + 6 x + 13 2 .

a) 5x + 10 = 5(x + 2)

25x2 + 50x = 25x(x + 2)

⇒ Nhân tử chung của chúng là: 5(x + 2)

b)

\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)

Ta có : E = 2x⁴ + 3x² + 7 

Mà : 2x⁴ \(\ge0\forall x\in R\)

        3x² \(\ge0\forall x\in R\)

Nên : E = 2x⁴ + 3x² + 7 \(\ge7\forall x\in R\)

Vây GTNN của E = 7 

Dấu "=" sảy ra khi : \(\hept{\begin{cases}2x^4=0\\3x^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^4=0\\x^2=0\end{cases}\Leftrightarrow}x=0}\)

3{x-2}²/x{x-2}{x²+2x+4}=3{x-2}/x²+2x+4=-3/x+2 

nho tick minh nha

\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)

\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)

\(a,3x-4y-3y+4x\)

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

\(=\left(3+4\right)\left(x-y\right)=7\left(x-y\right)\)

\(b,\left(a^3+2ab+b^2\right)-\left(a^3+b^3\right)\)

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

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

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

\(c,48b^3-24b^2=3b\)

\(48b^3-24b^2-3b=0\)

\(b\left(48b^2-24b-3\right)=0\)

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

\(81x^2y-27xy^2=27xy\left(3x-y\right)\)

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

\(25x^2-10xy+30xy^2=5x\left(5x-2y+6y^2\right)\)

\(27x^3-36xy=9x\left(3x^2-4y\right)\)

\(15x^2+30xy+25xy^2=5x\left(3x+6y+5y^2\right)\)

\(81x^2y-27xy^2==27xy\left(3x-y\right);2x^3+4x^2-6x=2x\left(x^2+2x-3\right)=2x\left(x+3\right)\left(x-1\right);25x^2-10xy+30xy^2=5x\left(5x-2y+6y^2\right);27x^3-36xy=3x\left(9x^2-12y\right);15x^2+30xy+25xy^2=5x\left(3x+6y+5y^2\right)\)