đầu tiên , x^4 + x^3 + 2X^2 +x+1 = (X^2)^2 + 2X^2 + 1 + X^3 + X = (x^2+1)^2 + x(X^2 +1) = ... đoạn này tự lm nha

Mình có cách khác :

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

Ai giúp mk vs

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

\(=x^2+6x+9-\left(2x-6x^2+6-18x\right)+9x^2+6x+1\)

\(=10x^2+12x+10+6x^2+16x-6=16x^2+28x+4\)

\(=4\left(4x^2+7x+1\right)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

(x² - x + 1)(x² + 2x -1)

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

=x(x²+2x+1)-3²

=x(x+1)²-3²

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

\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)