Question

Разложите на множители​

Answer 1

Ответ:

Пошаговое объяснение:

_______

Answer 2

Ответ$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\dddddd \displaystyle \Large \boldsymbol{:}$

Пошаговое объяснение:

$\displaystyle \large \boldsymbol{} 1) \ m^2-m-2=m^2-m-\underbrace {1-1}_{-2}=(m^2-1)-(m+1)=\\(m-1)(m+1)-(m+1)=(m-1-1)(m+1)=\boxed{(m-2)(m+1)} \\\\\\2) \ x^2-4x+3=x^2-\underbrace{x-3x}_{-4x}+3=x^2-x-(3x-3)=\\x(x-1)-3(x-1)=\boxed{(x-1)(x-3)} \\\\\\3) \ \ 16-(x^2+4x)^2=(4-x^2-4x)(\underbrace{x^2+4x+4}_{(x+2)^2})\\\\ -x^2-4x+4=0 \ |\cdot -1\\\\ . \ \ x^2+4x-4=0 \\\\D=16+16=32 \\\\x_1=\frac{4\sqrt{2} -4}{2}=\sqrt{8} -2 \ \ ; \ \ x_2=\frac{-4\sqrt{2}-4 }{2} =-\sqrt{8} -2$$\displaystyle \large \boldsymbol{} => -x^2-4x+4=-(x-\sqrt{8}+2 )(x+\sqrt{8}+2 )\\\\=> 16-(x^2+4x)^2=\boxed{-(x+2)^2 (x-\sqrt{8}+2 )(x+\sqrt{8}+2 )} \\\\\\4) \ \ x^3-y^6 =0 \ \ ; \ \ t=y^2 \ \ ; \ \ t^3=y^6 \ \ ; \ \ t^2=y^4\\\\ \ \ \ x^3-t^3=(x-t)(x^2+xt+t^2)\\\\x^3-y^6=\boxed{(x-y^2)(x^2+xy^2+y^4 )} \\\\\\5) \ x^3-7x+6=x^3-\underbrace{4x-3x}_{-7x}+6 =x(x^2-4)-(3x-6)= \\\\x\underline{(x-2)}(x+2)-3\underline{(x-2)}=(x-2)\underbrace{(x^2+2x-3)}_{(x+3)(x-1)}= \\\\\boxed{(x-2)(x-1)(x+3)}$

$\large \boldsymbol{} x^2+2x-3=x^2+2x-\underbrace{2-1}_{-3}=x^2-1+2(x-1)=\\(x+1)(x-1)+2(x-1)=(x+3)(x-1) \\\\\\6) \ x^3-13x-12= x^3-\underbrace{9x-4x}_{-13x}-12 = \\x^3-9x-(4x+12)=x(x+3)(x-3)-4(x+3)=\\\\ \underbrace{(x^2-3x-4)}_{(x-4)(x+1)}(x+3)=\boxed{ (x-4)(x+1)(x+3)}\\\\\\ x^2-3x-3-1=(x-1)(x+1)-3(x+1)=(x-4)(x+1)$