Thay x=-1 ta được -1^3+a.(-1)^2-2.a.(-1)+1=0

giải phương trình ta tìm được a=0

Giai pt (2) = 0 ta co 2 nghiem, x=1 va x=2

Thay lan luot vao phuong trinh 1 ta co he pt 2 an a va b, a=-3, b=0.

5x² - 4(x² - 2x + 1) - 5 = 0

=> 5x² - 4x² + 8x - 4 - 5 = 0 

=> x² + 8x - 9 = 0

=> x² + 9x - x - 9 = 0 

=> x(x + 9) - (x + 9) = 0

=> (x + 9)(x - 1) = 0

=> x + 9 = 0 => x = -9

hoặc x - 1 = 0 = > x = 1

                                                                       Vậy x = -9, x = 1

\(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\left(5x^2-5\right)-4\left(x^2-2.1.x+1^2\right)=0\)

\(5\left(x^2-1\right)-4\left(x-1\right)^2=0\)

\(5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)\left(x-1\right)=0\)

\(\left[5\left(x+1\right)-4\left(x-1\right)\right]\left(x-1\right)=0\)

\(\left(5x+5-4x+4\right)\left(x-1\right)=0\)

\(\left(x+9\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-9\\x=1\end{cases}}.\)

\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)

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

\(=\frac{\left(x+y-1\right)\left(x+y+1\right)}{\left(x-1-y\right)\left(x-1+y\right)}\)

\(=\frac{x+y+1}{x-y-1}\)

Gọi giá trị trên là : A

Ta có : \(A=x^2+x+2\)

\(=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{7}{2}\ge\frac{7}{2}\)

MAX \(A=\frac{7}{2}\Leftrightarrow x+\frac{1}{2}=0\Rightarrow x=\frac{-1}{2}\)

\(2x^2-x-15\)

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

\(x^4+x^2+1\)

a)   \(2x^2-x-15=2x^2-6x+5x-15=2x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(2x-5\right)\)

a)\(2x^2-x-15=2x^2-6x+5x-15\)

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

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

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

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

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

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