8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

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

4x² - 20x - [4x² - 3x - (4x - 3)] = 5

4x² - 20x - [4x² - 3x - 4x + 3] = 5

4x² - 20x - 4x_² + 3x + 4x - 3 = 5

-13x = 5 + 3

-13x = 8

x = \(\dfrac{\left(-8\right)}{13}\)

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

<=> 4x² - 20x - 4x² + 3x + 4x - 3 = 5

<=> 4x² - 4x² - 20x + 3x + 4x = 5 + 3

<=> -13x = 8

<=> \(x=-\dfrac{8}{13}\)

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

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, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

\(4x\left(3-\dfrac{1}{4}x\right)+\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow12x-x^2+x^2-4=0\Rightarrow12x=4\Rightarrow x=\dfrac{1}{3}\)

\(12x-x^2+x^2-2^2=0\)

\(12x-2=0\)

\(12x=2\)

\(x=\dfrac{1}{6}\)

Vậy x=1/6

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

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

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

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a)  \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)

\(x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)

Vậy .............................

b)  \(\left(2x-1\right)^2\)\(-\)\(\left(x+3\right)^2\)\(=\)\(0\)

\(\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)

\(\left(3x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)

Vậy ................................

c)  \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)

\(x^2\)\(\left(x-3\right)\)\(-\)\(4\)\(\left(x-3\right)\)\(=\)\(0\)

\(\left(x^2-4\right)\left(x-3\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)

a)\(x^3-\frac{1}{4}x=0\)

\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

b)\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{2}{3}\end{cases}}}\)