x² + 6x + 4ⁿ - 2ⁿ⁺¹ + 10 = 0

\(\Leftrightarrow\)( x² + 6x + 9 ) + ( 4ⁿ - 2ⁿ⁺¹ + 1 ) = 0

\(\Leftrightarrow\) ( x² + 2.3x + 3² ) + [(2ⁿ)² -2.2ⁿ + 1] = 0

\(\Leftrightarrow\) (x + 3)² + (2ⁿ - 1)² = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2^n-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\n=0\end{matrix}\right.\)

\(\Rightarrow\) x + n = -3

Ta có: x² + 6x + 4ⁿ - 2^n-1 +10 = 0 

\(\Rightarrow\) x² + 6x + 9 + 4ⁿ - 2^n-1 +1 = 0 

 \(\Rightarrow\)( x + 3)² + (2²)ⁿ - 2.2ⁿ +1 = 0

\(\Rightarrow\) ( x + 3)² + 2^2.n - 2  +2ⁿ +1 = 0

\(\Rightarrow\) \(\begin{cases}x+3=0\\2^n-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-3\\2^n=1\end{cases}\)\(\Leftrightarrow\begin{cases}x=-3\\n=0\end{cases}\)

\(\Rightarrow\)x + n = -3 +0 = -3

Chúc bạn học tốt 

b)\(x^4+6x^3+7x^2-6x+1=x^4+6x^3-2x^2+9x^2-6x+1\)

=\(x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)

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

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

c)\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

đặt \(x^2+10x+12=z\)

\(=\left(z-12\right)\left(z+12\right)+128=z^2-144+128\)

\(=z^2-16=\left(z-4\right)\left(z+4\right)\)\(=\left(x^2+10x-4+12\right)\left(x^2+10x+4+12\right)\)

\(=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(=\left(x^2+10x+8\right)\left(x^2+2x+8x+16\right)\)

\(=\left(x^2+10x+8\right)\left[x\left(x+2\right)+8\left(x+2\right)\right]\)

\(=\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)\)

a: =x^4-4x^3+x^2-3x^3+12x^2-3x+x^2-4x+1

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

c: =(x^2+10x)(x^2+10x+24)+128

=(x^2+10x)^2+24(x^2+10x)+128

=(x^2+10x+16)(x^2+10x+8)

=(x^2+10x+8)(x+2)(x+8)

a)

\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

\(=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+2x^2+6x^2+12x+5x+10}\)

\(=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+6x+5\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{x-2}{x+5}\)

b)

\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)

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

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

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

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((