9x² - 4 - ( 3x - 2 )( x + 5 ) = 0

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

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

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

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

x³ + 64 + ( x + 4 )( 2x - 3 ) = 0

<=> ( x + 4 )( x² - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0

<=> ( x + 4 )( x² - 4x + 16 + 2x - 3 ) = 0

<=> ( x + 4 )( x² - 2x + 13 ) = 0

<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x² - 2x + 13 = ( x² - 2x + 1 ) + 12 = ( x - 1 )² + 12 ≥ 12 > 0 ∀ x )

( x - 3 )( x² + 4x + 9 ) + 2( x² - 9 ) - 10( x - 3 ) = 0

<=> ( x - 3 )( x² + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0

<=> ( x - 3 )( x² + 4x + 9 + 2x + 6 - 10 ) = 0

<=> ( x - 3 )( x² + 6x + 5 ) = 0

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

<=> x = 3 hoặc x = -1 hoặc x = -5

<=> ( x - 3 )( 

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

Sos

a/ => x³ = 64 => x³ = 4³ => x = 4

b/ => 4x² - 12x + 9 - x² - 10x - 25 = 0 

=> 3x² - 22x - 16 = 0

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

=> x - 8 = 0 => x = 8

hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3

Vậy x = 8 ; x = -2/3

c/ => x³ - x² - 4x² + 8x - 4 = 0 

=> x³ - 5x² + 8x - 4 = 0 

=> (x - 2)² (x - 1) = 0

=> (x - 2)² = 0 => x - 2 = 0 => x = 2

hoặc x - 1 = 0 => x = 1 

Vậy x = 2 ; x = 1

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

a) Đặt t=x², t \(\ge0\)

Sau đó bạn giải theo tích ac

b) Dùng các hằng đẳng thức rồi phương pháp chuyển vế để giải PT

a: =>(x^2-4)(x^2-16)=0

=>x^2=4 hoặc x^2=16

=>\(x\in\left\{2;-2;4;-4\right\}\)

b: =>x^2-25-x^3-27=5-x^3+x^2+3x

=>-x^3+x^2-52=-x^3+x^2+3x+5

=>3x+5=-52

=>3x=-57

=>x=-19

\(\Leftrightarrow\left(x^4-20x^2+100\right)-36=0\)

\(\Leftrightarrow\left(x^2-10\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-10=6\\x^2-10=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=16\\x^2=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=\pm2\end{matrix}\right.\)

\(x^4-20x^2+64=0\)

Đặt \(t=x^2\)

\(PT\Leftrightarrow t^2-20t+64=0\\ \Leftrightarrow t^2-16t-4t+64=0\\ \Leftrightarrow t\left(t-16\right)-4\left(t-16\right)=0\\ \Leftrightarrow\left(t-16\right)\left(t-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t-16=0\\t-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}t=16\\t=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=16\\x^2=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{16}\\x\pm\sqrt{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm4\\x=\pm2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=2\\x=-2\end{matrix}\right.\\ Vậyx\in\left\{4;-4;2;-2\right\}\)