a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

\(x^4-9x^3+x^2-9x=0\)

\(\Leftrightarrow x\left(x^2+1\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

\(a,=x^2-4-x^2+2x+4=2x\\ b,=\left(x-5y\right)^2:\left(5y-x\right)=\left(5y-x\right)^2:\left(5y-x\right)=5y-x\\ c,Sửa:\left(28x-9x^2+x^3-30\right):\left(x-3\right)\\ =\left(x^3-3x^2-6x^2+18x+10x-30\right):\left(x-3\right)\\ =\left(x-3\right)\left(x^2-6x+10\right)\left(x-3\right)=x^2-6x+10\)

Thu gọn x5 - 3x2 + 7x4 - 9x3 + x2 - 2x + 5x2 = x5 + 7x4 - 9x3 + 3x2 - 2x

Hệ số cao nhất là 1, hệ số tự do là 0. Chọn A

\(4x-9.3x-10.2x=0\)

\(=>4x-27x-20x=0\)

\(=>x\left(4-27-20\right)=0\)

\(=>x.\left(-43\right)=0\)

\(=>x=0\)

P(x)=-31x^3+4x^4+4x^2+15=4x^4-31x^3+4x^2+15

P(1)=4-31+4+15=23-31=-8

P(0)=15

P(-1)=4+31+4+15=56

lm đuy:)_{vừa nhạc hay thế còn j lại chê}

\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)

a: Đặt M(x)=0

=>\(2x-\frac12=0\)

=>\(2x=\frac12\)

=>\(x=\frac14\)

b: Đặt N(x)=0

=>\(\left(x+5\right)\left(4x^2-1\right)=0\)

=>(x+5)(2x-1)(2x+1)=0

=>\(\left[\begin{array}{l}x+5=0\\ 2x-1=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=\frac12\\ x=-\frac12\end{array}\right.\)

c: Đặt P(x)=0

=>\(x\left(9x^2-25\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ 9x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ 9x^2=25\end{array}\right.\)

=>\(\left[\begin{array}{l}x=0\\ x^2=\frac{25}{9}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac53\\ x=-\frac53\end{array}\right.\)