\(\Leftrightarrow\dfrac{\left(x^2+5\right)\left(x^2-5\right)+2x\left(x^2+5\right)}{x^2+5}=3\)

=>x^2+2x-5=3

=>x^2+2x-8=0

=>(x+4)(x-2)=0

=>x=-4 hoặc x=2

a: \(x^2-y^2-x-y\)

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

\(=\left(x+y\right)\left(x-y-1\right)\)

f: \(x^3-5x^2-5x+1\)

\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)

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

a. 5x + 3(x² - x - 1)

= 5x + 3x² - 3x - 3

= 3x² + 5x - 3x - 3

= 3x² + 2x - 3

b. (5 - x)(5 + x) - (2x - 1)²

25 - x² - (4x² - 4x + 1)

= 25 - x² - 4x² + 4x - 1

= 25 - 1 - x² - 4x² + 4x 

= 24 - 5x² + 4x

a) \(5x+3\left(x^2-x-1\right)=5x+3x^2-3x-3=3x^2+2x-3\)

b) \(\left(5-x\right)\left(5+x\right)-\left(2x-1\right)^2=25-x^2-4x^2+4x-1=-5x^2+4x+24\)

\(a) 5x+3(x^2-x-1)=5x+3x^2-3x-3\)

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

\(b)(5-x)(5+x)-(2x-1)^2=25-x^2-4x^2+4x-1\)

\(=-5x^2+4x-24\)

a: \(=\dfrac{\left(x^2+5\right)\left(x^2-5\right)+2x\left(x^2+5\right)}{x^2+5}=x^2+2x-5\)

b: \(=\dfrac{x^3-2x^2-x^2+2x+3x-6}{x-2}=x^2-x+3\)

\(g,x^4-16=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ i,-x^2+10x-25=-\left(x-5\right)^2\\ k,x^3+3x^2+3x+1-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\\ =\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ m,\left(x+y\right)^2-25\left(x+y\right)+24=\left(x+y-5\right)^2-1=\left(x+y-4\right)\left(x+y-6\right)\)

g. x⁴ - 16

<=> x⁴ - 4²

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

i. -x² + 10x - 25

<=> -(x² - 10x + 25)

<=> -(x² -10x + 5²)

<=> -(x - 5)²

sao ko ai like tui, buồn

a) \(N=x^2-10x+25\)

\(N=x^2-2\cdot5\cdot x+5^2\)

\(N=\left(x-5\right)^2\)

Thay x = 55 vào N ta có: 

\(N=\left(55-5\right)^2=2500\)

b) \(P=\dfrac{x^4}{4}-x^2y+y^2\)

\(P=\left(\dfrac{x^2}{2}\right)^2-2\cdot\dfrac{x^2}{2}\cdot y+y^2\)

\(P=\left(\dfrac{x^2}{2}-y\right)^2\)

Thay x = 4 và \(y=\dfrac{1}{2}\) vào P ta có:

\(P=\left(\dfrac{4^2}{2}-\dfrac{1}{2}\right)^2=\dfrac{225}{4}\)

a)\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)

\(\Rightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2=x^2+6x+64\)

\(\Rightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)

\(\Rightarrow8^2=x^2+6x+64\)

\(\Rightarrow64=x^2+6x+64\)

\(\Rightarrow x^2+6x=0\)

\(\Rightarrow x\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

b) \(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)=3\)

\(\Rightarrow\left(x^4+5x^2-5x^2-25+2x^3+10x\right):\left(x^2+5\right)=3\)

\(\Rightarrow\left[x^2\left(x^2+5\right)-5\left(x^2+5\right)+2x\left(x^2+5\right)\right]:\left(x^2+5\right)=3\)

\(\Rightarrow\left(x^2+5\right)\left(x^2-5+2x\right):\left(x^2+5\right)=3\)

\(\Rightarrow x^2+2x-5=3\)

\(\Rightarrow x^2+2x-5-3=0\)

\(\Rightarrow x^2+2x-8=0\)

\(\Rightarrow x^2+4x-2x-8=0\)

\(\Rightarrow x\left(x+4\right)-2\left(x+4\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Mình cần gấp ạk

A và B