tìm y nữa 

mình viết thiếu

a. (5x-1)²  -  (5x-4) (5x-4) +7

= (5x-1)² - (5x-4)²  + 7

=[(5x-1)+(5x-4)] [(5x-1)-(5x-4)] +7  ( đoạn này bỏ cx đc)

=(10x-5) .3+7

=30x-15+7

=30x-8

ý a hơi sai sai

b: \(8x^2-48x+6xy-36y\)

\(=8x\left(x-6\right)+6y\left(x-6\right)\)

\(=2\left(x-6\right)\left(4x+3y\right)\)

d: \(a^2-2ab+b^2-4\)

\(=\left(a-b\right)^2-4\)

\(=\left(a-b-2\right)\left(a-b+2\right)\)

\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)

\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

a: \(\left(2x^4-x^3+5x^{}-6x^2-1\right):\left(1-2x\right)\)

\(=\frac{2x^4-x^3-6x^2+5x-1}{-2x+1}\)

\(=\frac{-2x^4+x^3+6x^2-5x+1}{2x-1}=\frac{-x^3\left(2x-1\right)+6x^2-3x-2x+1}{2x-1}\)

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

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

\(=\frac{x^3-x^2+x-2x^2+2x-2}{x^2-x+1}\)

\(=\frac{x\left(x^2-x+1\right)-2\left(x^2-x+1\right)}{x^2-x+1}=x-2\)

c: \(\left(2x^3+5x^2-2x+3\right):\left(2x^2-x+1\right)\)

\(=\frac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}\)

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

d: \(\frac{5xy^2}{x^2y}+\frac{5y-x^2}{xy^2}\)

\(=\frac{y\cdot5xy^2+x\left(5y-x^2\right)}{x^2y^2}=\frac{5x^2y^2+5xy-x^3}{x^2y^2}=\frac{x\left(5xy^2+5y-x^2\right)}{x^2y^2}\)

\(=\frac{5xy^2+5y-x^2}{xy^2}\)

e: \(\frac{27-x^3}{5x^5}:\frac{2x-6}{3x^3}\)

\(=\frac{-\left(x-3\right)\left(x^2+3x+9\right)}{5x^5}\cdot\frac{3x^3}{2\left(x-3\right)}=\frac{-3\left(x^2+3x+9\right)}{5x^2\cdot2}=\frac{-3\left(x^2+3x+9\right)}{10x^2}\)

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)

$\textbf{Bài 1d}$

$x^2-2|x-3|+3x\ge0.$

$\bullet\ x\ge3:$

$x^2-2(x-3)+3x\ge0$

$\Leftrightarrow x^2+x+6\ge0.$

Bất phương trình luôn đúng.

$\bullet\ x<3:$

$x^2-2(3-x)+3x\ge0$

$\Leftrightarrow x^2+5x-6\ge0$

$\Leftrightarrow(x+6)(x-1)\ge0.$

$\Leftrightarrow x\le-6$ hoặc $x\ge1.$

Kết hợp điều kiện $x<3$.

Vậy $x\le-6$ hoặc $1\le x.$