Sửa đề: 2016x^2

x^4+2016x^2+2015x+2016

=x^4+x^3+x^2-x^3-x^2-x+2016x^2+2016x+2016

=(x^2+x+1)(x^2-x+2016)

2015x⁴ + 2016x² + x + 2016

= (2015x⁴ + 2015x³ + 2015x²) + (- 2015x³ - 2015x² - 2015x) + (2016x² + 2016x + 2016)

= (x² + x + 1)(2015x² - 2015x + 2016)

Vào câu trả lời tương tự đi có đáp án đó

Ta có: x^4 + 2016x^2 + 2015x + 2016

= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2016x^2 + 2016x + 2016

= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2016(x^2 + x + 1)

= (x^2 + x + 1)(x^2 - x + 2016)

       \(x^4+2016x^2+2015x+2016\)

\(=x^4-x+2016x^2+2016x+2016\)

\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

 x⁴+2015x²+2014x+2015

=x⁴-x+2015x²+2015x+2015

=x.(x³-1)+2015.(x²+x+1)

=x.(x-1)(x²+x+1)+2015.(x²+x+1)

=(x²+x+1)(x²-x+2015)

\(x^4+2015x^2+2014x+2015=\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(2015x^2+2015x+2015\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)

\(x^4+2015x^2+2014x+2015.\)

=\(\left(x^4-x\right)+2015x^2+2015x+2015\)

=\(x\left(x^3-1\right)+2015\left(x^2+x+1\right)\)

=\(x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left(x^2-x-2015\right)\)

k cho mik

trả lời

xx^4+2015x^2+2014x+2015=x^4+2015x^2+2015x-x+2015=x\left(x^3-1\right)+2015\left(X^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)xx4+2015x2+2014x+2015=x4+2015x2+2015x−x+2015=x(x3−1)+2015(X2+x+1)=x(x−1)(x2+x+1)+2015(x2+x+1)=(x2+x+1)(x2−x+2015)

$hc\; tốt$

\(x^4+2015x^2+2014x+2015\)

\(=\left(x^4-x\right)+2015x^2+2015x+2015\)

\(=x\left(x^3-1\right)+2015\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)

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

x⁴+4 = (x²)²+2² = x⁴ + 2.x².2 + 4 – 4x²

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

Ta có: \(x^4+4\)

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

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

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

x⁴ + x²y² +y⁴                   

= (x²)² +  x²y² + (y²)²          

= (x²)² +  x²y² + (y²)²  + x²y² - x²y²       

= (x²)² +  2 x²y² + (y²)²  - x²y²      

= (x² + y²)²- (xy)²                  

=(x² + y² + xy)(x² + y² - xy)