b: 4x^2-20x+25=(x-3)^2

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

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

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

=>(3x-8)(x-2)=0

=>x=8/3 hoặc x=2

c: x+x^2-x^3-x^4=0

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

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

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

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

=>\(x\in\left\{0;1;-1\right\}\)

d: 2x^3+3x^2+2x+3=0

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

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

=>2x+3=0

=>x=-3/2

a: =>x^2(5x-7)-3(5x-7)=0

=>(5x-7)(x^2-3)=0

=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)

a. `4x^2-20x+25=0`

`<=>(2x)^2-2.2x.5 +5^2=0`

`<=>(2x-5)^2=0`

`<=>2x-5=0`

`<=>x=5/2`

b. `(x-5)(x+5)-(x-3)^2=2(x-7)`

`<=>x^2-25-x^2+6x-9=2x-14`

`<=>6x-34=2x-14`

`<=>4x=20`

`<=>x=5`

\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)

\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)

\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)

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

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

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

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

=>\(\left[\begin{array}{l}x+5=0\\ 3x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=-\frac13\end{array}\right.\)

b: \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)

=>\(\left\lbrack3\left(2x+1\right)\right\rbrack^2-\left\lbrack2\left(x+1\right)\right\rbrack^2=0\)

=>\(\left(6x+3\right)^2-\left(2x+2\right)^2=0\)

=>(6x+3+2x+2)(6x+3-2x-2)=0

=>(8x+5)(4x+1)=0

=>\(\left[\begin{array}{l}8x+5=0\\ 4x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac58\\ x=-\frac14\end{array}\right.\)

c: \(x^3-6x^2+9x=0\)

=>\(x\left(x^2-6x+9\right)=0\)

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

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

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

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

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

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

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

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

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

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

=>-4(x-2)=0

=>x-2=0

=>x=2

g: \(x^4-2x^2+1=0\)

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

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

=>\(x^2=1\)

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

h: \(4x^2+y^2-20x-2y+26=0\)

=>\(4x^2-20x+25+y^2-2y+1=0\)

=>\(\left(2x-5\right)^2+\left(y-1\right)^2=0\)

=>\(\begin{cases}2x-5=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=\frac52\\ y=1\end{cases}\)

i: \(x^2-2x+5+y^2-4y=0\)

=>\(x^2-2x+1+y^2-4y+4=0\)

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

=>\(\begin{cases}x-1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=2\end{cases}\)

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

\(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

x ∈ N*

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

`(4x^2-3)^2=-8`

Vì `(4x^2-3)^2 >=0> -8` với mọi `x` nên PT trên vô nghiệm.

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

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow x\left(x-2\right)=0\) (do \(x^2+10>0;\forall x\))

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

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

`x^4-2x^3+10x^2-20x=0`

`<=>x^3(x-2)+10x(x-2)=0`

`<=>(x^3+10x)(x-2)=0`

`<=>x(x^2+10)(x-2)=0`

`<=>`$\left[\begin{matrix} x=0\\ x^2+10=0\\x-2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=0\\ x^2=-10 \ \rm(loại) \\x=2\end{matrix}\right.$

Vậy `S={0;2}`

( 2 x   –   3 ) 2   –   4 x 2   +   9   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( 4 x 2   –   9 )   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( ( 2 x ) 2   –   3 2 )   =   0     ⇔   ( 2 x   –   3 ) 2   –   ( 2 x   –   3 ) ( 2 x   +   3 )   =   0

ó (2x – 3)(2x – 3 – 2x – 3) = 0

ó (2x – 3)(-6) = 0

ó 2x – 3 = 0

ó x = 3 2

Đáp án cần chọn là: C