\(x\left(5-6x\right)+\left(2x-1\right)\left(3x+\text{4}\right)=6\\ \Leftrightarrow5x-6x^2+6x^2+8x-3x-4=6\)

\(\Leftrightarrow10x-4=6\)

\(\Leftrightarrow10x=6+4\\ \Leftrightarrow10x=10\\ \Leftrightarrow x=\dfrac{10}{10}\)

\(\Leftrightarrow x=1\)

\(x^2\left(x-2021\right)-x+2021=0\)

\(\Leftrightarrow x^2\left(x-2021\right)-(x-2021)=0\)

\(\Leftrightarrow\left(x-2021\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-2021\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2021=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=1\\x=-1\end{matrix}\right.\)

a) \(x^2+3x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)

a: Ta có: \(x^2+3x-4=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

b: Ta có: \(x^2-2x-1=0\)

\(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(-1\right)=8\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2-2\sqrt{2}}{2}=1-\sqrt{2}\\x_2=\dfrac{2+2\sqrt{2}}{2}=1+\sqrt{2}\end{matrix}\right.\)

a) \(2x\left(x+4\right)-\left(x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2+8x-2x^2-x+3=0\)

\(\Leftrightarrow7x=-3\Leftrightarrow x=-\dfrac{3}{7}\)

b) \(x^2-2x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(a,\Leftrightarrow2x^2+8x-2x^2-x+3=0\\ \Leftrightarrow7x=-3\\ \Leftrightarrow x=-\dfrac{3}{7}\\ b,x^2-2x-3=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

a: Ta có: \(2x\left(x+4\right)-\left(x-1\right)\cdot\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2+8x-2x^2-3x+2x+3=0\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

b: ta có: \(x^2-2x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

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 \(x=\dfrac{5}{4}\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

a: =>2x^2=4

=>x^2=2

=>\(x=\pm\sqrt{2}\)

b: =>(x+1)^2-4=0

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

=>(x+3)(x-1)=0

=>x=1 hoặc x=-3

c: =>(2x-1)^2-3^2=0

=>(2x-1-3)(2x-1+3)=0

=>(2x-4)(2x+2)=0

=>x=2 hoặc x=-1

d: x^2-x=0

=>x(x-1)=0

=>x=0 hoặc x=1

Bài 2: 

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

=>-13x=26

hay x=-2

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

hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)

c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

hay \(x\in\left\{-5;2\right\}\)

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

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

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)