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

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

\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)

a) x² - 4 = 0

x² = 4

x = 2 hoặc x = -2

b) 2x(x + 5) - 3(5 + x) = 0

(x + 5)(2x - 3) = 0

X + 5 = 0 hoặc 2x - 3 = 0

*) x + 5 = 0

x = -5

*) 2x - 3 = 0

2x = 3

x = 3/2

c) x³ - 6x² + 11x - 6 = 0

x³ - x² - 5x² + 5x + 6x - 6 = 0

(x³ - x²) - (5x² - 5x) + (6x - 6) = 0

x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

(x - 1)(x² - 5x + 6) = 0

(x - 1)(x² - 2x - 3x + 6) = 0

(x - 1)[(x² - 2x) - (3x - 6)] = 0

(x - 1)[x(x - 2) - 3(x - 2)] = 0

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

x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x - 3 = 0

x = 3

Vậy x = 1; x = 2; x = 3

a: \(2x^3-50x=0\)

=>\(2x\left(x^2-25\right)=0\)

=>x(x-5)(x+5)=0

=>x∈{0;5;-5}

b: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

=>2x(3x-5)+(3x-5)=0

=>(3x-5)(2x+1)=0

=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)

c: \(9\left(3x-2\right)=x\left(2-3x\right)\)

=>9(3x-2)-x(2-3x)=0

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

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

=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)

d: \(\left(2x-1\right)^2-25=0\)

=>(2x-1-5)(2x-1+5)=0

=>(2x-6)(2x+4)=0

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

=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)

e: \(25x^2-2=0\)

=>\(25x^2=2\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

f: \(x^2-25=6x-9\)

=>\(x^2-6x-16=0\)

=>(x-8)(x+2)=0

=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)

g: 5x(x-3)-2x+6=0

=>5x(x-3)-2(x-3)=0

=>(x-3)(5x-2)=0

=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)

h: 3x(x-7)-2(x-7)=0

=>(x-7)(3x-2)=0

=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)

i: \(7x^2-28=0\)

=>\(7x^2=28\)

=>\(x^2=4\)

=>x=2 hoặc x=-2

j: 2x+1+x(2x+1)=0

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

=>\(\left[\begin{array}{l}2x+1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-1\end{array}\right.\)

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

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

=>4(x+2)=0

=>x+2=0

=>x=-2

l: \(x^3+5x^2-4x-20=0\)

=>\(x^2\left(x+5\right)-4\left(x+5\right)=0\)

=>\(\left(x+5\right)\left(x^2-4\right)=0\)

=>(x+5)(x-2)(x+2)=0

=>x∈{-5;2;-2}

m: \(x^2-25+2\left(x+5\right)=0\)

=>(x-5)(x+5)+2(x+5)=0

=>(x+5)(x-3)=0

=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=3\end{array}\right.\)

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

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

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

o: \(x^2-6x+8=0\)

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

=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)

p: \(x^2-5x-14=0\)

=>\(x^2-7x+2x-14=0\)

=>(x-7)(x+2)=0

=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)

q: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

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

=>-4x+13=6

=>-4x=6-13=-7

=>x=7/4

r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)

=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)

=>-4x+26=18