a) 1/2(x³+8)=1/2(x+2)(x²-2x+4)

b) x⁴(x-y)+2x³(x-y)=x³(x+2)(x-y)

c) x²-(y²-6y+9)=x²-(y-3)²=(x-y+3)(x+y-3)

d) xy(x³+y³)=xy(x+y)(x²-xy+y²)

e)3x²(x²-25y²)=3x²(x-5y)(x+5y)

f) 4x⁴+4x²y²+y⁴-4x²y²= (2x²+y²)²-(2xy)²=(2x²-2xy+y²)(2x²+2xy+y²)

a) \(\frac{1}{2}x^3+4=\frac{1}{2}\left(x^3+8\right)=\frac{1}{2}\left(x+2\right)\left(x^2-2x+4\right)\)

b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)

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

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

e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).

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

Lời giải:

a.

$A=-5x^4+\frac{2}{5}x^2y+y^4-\frac{2}{5}x^2y=-5x^4+y^4$

Bậc của A: $4$

b.

ĐKXĐ của phân thức: $x+2\neq 0\Leftrightarrow x\neq -2$
 

thank bạn

a) ĐKXĐ: \(x^2+6x+11\ge0\)đúng\(\forall x\inℝ\)

b) ĐKXĐ: \(\hept{\begin{cases}\left(2x-3\right)\left(x+2\right)\ge0\\x+3\ne0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\le-2,x\ne-3\\x\ge\frac{3}{2}\end{cases}}}\)

c) ĐKXĐ: \(-x^2-5\ge0\)Vô nghiệm\(\forall x\inℝ\)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)

a) điều kiện xác định: x≠3 và x≠2

b) \(\dfrac{x^2-4}{\left(x-3\right)\left(x-2\right)}\)=\(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x-2\right)}\)=\(\dfrac{x+2}{x-3}\)

Tại x=13 ta có \(\dfrac{13+2}{13-3}\)=\(\dfrac{3}{2}\)

a, ĐKXĐ: x\(\ne\)5, x\(\ne\)0, x\(\ne\)-5

b, B = \(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

     = \(\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

     =\(\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2x^2-50}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

    = \(\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

    =\(\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)=\(\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)=\(\frac{x-1}{2}\)

Với B = 0 thì\(\frac{x-1}{2}\)=0 => x = 1

Với B = \(\frac{1}{4}\)thì \(\frac{x-1}{2}\)=\(\frac{1}{4}\)=> x = 1,5

`a,x^3-8 ne 0`

`=>x^3 ne 8`

`=>x ne 2`

`b,2x^2+5x+3 ne 0`

`=>2x^2+2x+3x+3 ne 0`

`=>2x(x+1)+3(x+1) ne 0`

`=>(x+1)(2x+3) ne 0`

`=>x ne -1,-3/2`

`c,x^2-4 ne 0`

`=>x^2 ne 4`

`=>x ne 2,-2`

a) ĐK:

 \(x^3-8\ne0\\ \Leftrightarrow x\ne2\)

b) ĐK:

 \(2x^2+5x+3\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne-\dfrac{3}{2}\end{matrix}\right.\)

c) ĐK:

\(x^2-4\ne0\\ \Leftrightarrow x\ne\pm2\)

a) ĐKXĐ: \(x\ne2\)

b) ĐKXĐ: \(x\notin\left\{-\dfrac{3}{2};-1\right\}\)

c) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)