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

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

=>x(x-15)=0

=>x=0 hoặc x-15=0

=>x=0 hoặc x=15

\(x^2-15x=0\)

\(\Leftrightarrow x\left(x-15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=15\end{matrix}\right.\)

Vậy \(S=\left\{0;15\right\}\)

a) Ta có: \(7x\left(x-20\right)-x+20=0\)

\(\Leftrightarrow\left(x-20\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=\dfrac{1}{7}\end{matrix}\right.\)

b) Ta có: \(x^3-15x=0\)

\(\Leftrightarrow x\left(x-\sqrt{15}\right)\left(x+\sqrt{15}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)

   15x-15=0
=>15x    =0+15
=>15x    =15
=>   x    =15:15
=>   x    =  1

|15x-15|=0

nên 15x-15=0

15x=0+15

15x=15

x=15/15

x=1

Vậy x=1

Chúc bạn học tốt!^_^

Ta có

x 3   +   3 x 2   +   3 x   +   1   =   0   ⇔   ( x   +   1 ) 3   =   0

ó x + 1 = 0 ó x = -1

Vậy x = -1

Đáp án cần chọn là: A

b) Ta có: 5x3 – 3x2 + 10x – 6 = (5x3 + 10x )+ ( -3x2– 6)

= 5x(x2 + 2) – 3(x2 + 2) = (x2 + 2)(5x – 3)

Vậy (x2 + 2)(5x – 3) = 0 ⇒ 5x – 3 = 0 (vì x2 + 2 ≥ 0, với mọi x)

⇒x = 3/5

nha nha\

\(\frac{1}{6x}+\frac{1}{10x}-\frac{4}{15x}+1=0\)

\(\Rightarrow\frac{1}{x}\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(\Rightarrow\frac{1}{x}.0+1=0\)

\(\Rightarrow1=0\)( vô lý )

=> không có giá trị x thỏa mãn đề bài 

Study well 

\(\frac{1}{6x}+\frac{1}{10x}-\frac{4}{15x}+1=0\)

\(\Rightarrow\frac{1}{12}\times\frac{2}{x}+\frac{1}{20}\times\frac{2}{x}-\frac{2}{15}\times\frac{2}{x}=-1\)

\(\Rightarrow\frac{2}{x}\times\left(\frac{1}{12}+\frac{1}{20}-\frac{2}{15}\right)=-1\)

\(\Rightarrow\frac{2}{x}\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}-\frac{2}{3\times5}\right)=-1\)

\(\Rightarrow\frac{2}{x}\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}-\frac{1}{3}-\frac{1}{5}\right)=-1\)

\(\Rightarrow\frac{2}{x}\times0=-1\)

\(\Rightarrow x\in\varnothing\)