Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

a) x = 8 3 .                            b) x = − 9 20 .  

a) \(3x^3-6x^2=0\)

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

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

\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b) \(x\left(x-4\right)-12x+48=0\)

\(x^2-4x-12x+48=0\)

\(x^2-16x+48=0\)

\(\left(x-12\right)\left(x-4\right)=0\)

\(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)

\(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)

c) Viết thiếu nha :v

d) \(2x\left(x-5\right)-x\left(2x+3\right)=16\)

\(2x^2-10x-x^2-2x^2-3x=16\)

\(-13x=16\)

\(x=-\frac{16}{13}\)

e) \(\left(4x^2-1\right)-\left(x-1\right)^2=-3\)

\(4x^2-1-x^2+2x-1=-3\)

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

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

\(3x^2+1+2x=0\)

Vì \(3x^2+1+2x>0\)nên: 

\(x\in\varnothing\)

A) 3x³ - 6x² = 0

=> 3x²(x - 2) = 0

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

=> \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b) x(x - 4) - 12x + 48 = 0

=> x(x - 4) - 12(x - 4) = 0

=> (x - 12)(x - 4) = 0

=> \(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)

c) x(x - 4) - (x² - 8) = x² - 4x - x² + 8 = 4x + 8 

\(a,3x^3-6x^2=0\Rightarrow3x^2\left(x-2\right)=0.\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

\(b,x\left(x-4\right)-12x+48=0\)

\(\Rightarrow x\left(x-4\right)-12\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-12=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=12\end{cases}}}\)

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

\(\Rightarrow c,x^2-4x-x^2+8=0\)

\(\Rightarrow-4x+8=0\)

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

\(\Rightarrow x=2\)

\(d,2x\left(x-5\right)-x\left(2x+3\right)=16\)

\(\Rightarrow2x^2-10x-2x^2-3x=16\)

\(\Rightarrow-13x=16\Leftrightarrow x=-\frac{16}{13}\)

\(e,\left(4x^2-1\right)-\left(x-1\right)^2=-3\)

\(\Rightarrow4x^2-1-x^2+2x-1=-3\)

\(\Rightarrow3x^2+2x+1=0\)

\(\Rightarrow x^2+\frac{2}{3}x+\frac{1}{3}=0\)

\(\Rightarrow x^2+2.x.\frac{1}{3}+\frac{1}{9}+\frac{2}{9}=0\)

\(\Rightarrow\left(x+\frac{1}{3}\right)^2+\frac{2}{9}=0\)( vô lý ) 

Vậy phương trình vô nghiệm

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 \(x\in R\)

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

Bài 1.

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a. \(3x^3-6x^2=0\Leftrightarrow3x^2\left(x-2\right)=0\)

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

b. \(x\left(x-4\right)-12x+48=0\)

\(\Leftrightarrow x\left(x-4\right)-12\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=12\end{matrix}\right.\)

V.v.v.v

a) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.