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

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

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

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

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

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

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

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

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

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

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

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

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

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

b: \(=\dfrac{2x^3-4x^2+3x^2-6x+x-2}{x-2}=2x^2+3x+1\)

d: \(=\dfrac{x^4-2x^3+3x^2+2x^3-4x^2+6x-x^2+2x-3}{x^2-2x+3}=x^2+2x-1\)

\(x^4-x^2+10x-25\)

\(=x^4-\left(x^2-10x+25\right)\)

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

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

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

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

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

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

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

b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)

\(=2x^2-3xy+5y^2\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

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

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

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

\(\Leftrightarrow A=x^2+4xy+3x^2+x^2+y^2-4xy\)

\(\Leftrightarrow A=5x^2+y^2\)

\(B+\left(-x^4+x^2-2x^3-\dfrac{1}{3}\right)=3x^2-2x^3+x-\dfrac{2}{3}\)

\(\Leftrightarrow B=3x^2-2x^3+x-\dfrac{2}{3}+x^4-x^2+2x^3+\dfrac{1}{3}\)

\(\Leftrightarrow B=x^4+2x^2+x-\dfrac{1}{3}\)

A=5x²+y².

B=x⁴+2x²+x-1/3.