\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

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

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

a) \(3x+20:x+5\)\(=3\left(x+5\right)+5:x+5\)

Vậy đế \(3x+20⋮x+5\) thì \(x+5\inƯ\left(5\right)\)

Mà Ư(5)={1;-1;5;-1}

 =>x+5={1;-1;5;-5}

Ta có bảng sau:

Vậy x={-4;-6;0;-10}

a/ Ta có \(3x+20=3\left(x+5\right)+5\)

Để \(3x+20⋮\left(x+5\right)\) thì \(5⋮\left(x+5\right)\)

Xét các trường hợp

b,c tương tự

a) Ta có:

\(3x+20⋮x+5\)

\(\Rightarrow\left(3x+15\right)+5⋮x+5\)

\(\Rightarrow3\left(x+5\right)+5⋮x+5\)

\(\Rightarrow5⋮x+5\)

\(\Rightarrow x+5\in\left\{\pm1;\pm5\right\}\)

+) \(x+5=1\Rightarrow x=-4\)

+) \(x+5=-1\Rightarrow x=-6\)

+) \(x+5=5\Rightarrow x=0\)

+) \(x+5=-5\Rightarrow x=-10\)

Vậy \(x\in\left\{-4;-6;0;-10\right\}\)

a) \(\left(x-1\right)^3\)

\(=x^3-3x^2+3x-1\)

b) \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

Bài 3: 

a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)

\(\Leftrightarrow12x=13\)

hay \(x=\dfrac{13}{12}\)

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)

\(\Leftrightarrow x^3-1-x^3+4x=4\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

a) \(3x\cdot\left(12x-4\right)-9x\cdot\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=\dfrac{30}{15}\)

\(\Leftrightarrow x=2\)

b) \(\left(2x+1\right)-5\left(x-2\right)=10\)

\(\Leftrightarrow2x+1-5x+10=10\)

\(\Leftrightarrow-3x+11=10\)

\(\Leftrightarrow-3x=10-11\)

\(\Leftrightarrow-3x=-1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(P\left(x\right)-Q\left(x\right)=7x^2-9x+6\)

Để TMĐK đề bài thì: \(7x^2-9x+6=2x^2-3x+6\)

\(\Leftrightarrow5x^2-6x=0\Leftrightarrow x\left(5x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{5}\end{cases}}\)

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

\(Q\left(x\right)=-2x^3+2x^2+4x+15\)

\(P\left(x\right)+Q\left(x\right)=-4x^3+5x^2+8x+21\)

\(P\left(x\right)=Q\left(x\right)\)

\(\Leftrightarrow-2x^3+3x^2+4x+6=-2x^3+2x^2+4x+15\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

a) -4x^3+5x^2+8x+21

b: \(\Leftrightarrow\dfrac{x-2}{A}=\dfrac{\left(5x-1\right)\left(x-2\right)}{x^2\left(5x-1\right)+3\left(5x-1\right)}=\dfrac{x-2}{x^2+3}\)

hay \(A=x^2+3\)