\(x^5+x+1\)

\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

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

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

Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.

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

Câu 1:

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

\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)

M = x⁹ - x⁷ + x⁶ - x⁵ - x⁴ + x³ - x² + 1

= ( x⁹ - x⁷ ) + ( x⁶ - x⁴ ) - ( x⁵ - x³ ) - ( x² - 1 )

= x⁷( x² - 1 ) + x⁴( x² - 1 ) - x³( x² - 1 ) - ( x² - 1 )

= ( x² - 1 )( x⁷ + x⁴ - x³ - 1 )

= ( x - 1 )( x + 1 )[ x⁴( x³ + 1 ) - ( x³ + 1 ) ]

= ( x - 1 )( x + 1 )( x³ + 1 )( x⁴ - 1 )

= ( x - 1 )( x + 1 )( x + 1 )( x² - x + 1 )( x² - 1 )( x² + 1 )

= ( x + 1 )²( x - 1 )( x² - x + 1 )( x - 1 )( x + 1 )( x² + 1 )

= ( x + 1 )³( x - 1 )²( x² + 1 )( x² - x + 1 )

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

= (x^3 - x) + (x^2y - y)

= x(x^2 - 1) + y(x^2 - 1)

= ( x^2 -1)(x+y)

Ta có:

\(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)

\(=\left(x^9-x^8\right)+\left(x^8-x^7\right)-\left(x^6-x^5\right)-\left(2x^5-2x^4\right)-\left(x^4-x^3\right)+\left(x^2-x\right)+\left(x-1\right) \)

\(=x^8.\left(x-1\right)+x^7.\left(x-1\right)-x^5.\left(x-1\right)-2x^4.\left(x-1\right)-x^3\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left(x^8+x^7-x^5-2x^4-x^3+x+1\right)\)

xin chào làm ơn đừng trách mk mk sẽ nói cách giải

nguyễn bích hằng sao lại trách

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