g) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

  \(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

  \(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

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

i) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

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

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

a, 3x2 - 8x2 - 2x+3=0

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

2x5 - 2x+3=0

2x5 - 2x=0-3=

2x5 - 2x=-3

2x(5-x)=-3

5-x=-3/2

5-x=1,5

x=5-1,5

x=3,5

3,5 nha bn

chúc bn học tốt

happy new year

ảnh lỗi r ạ

Lỗi rùi

lưu ảnh rồi gửi vào đây nhé

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

a) 4x^2 - 25 - ( 2x - 5) .( 2x + 7) = 0

<=>4x²-25-(4x²+14x-10x-35)=0

<=>4x²-25-4x²-14x+10x+35= 0 

<=>-4x+10= 0 

<=>x= 5/2

b)  x^3 + 27 + ( x+3). ( x -9) = 0

<=>x³+3³+(x+3)(x-9)=0

<=>(x+3)(x²-3x+9)+(x+3)(x-9)=0

<=>(x+3)(x²-3x+9+x-9) =0

<=>(x+3)(x²-2x)=0

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

<=>x=-3 hoặc x=2 hoặc x=2