`a)16x^2-24x+9=25`

`<=>(4x-3)^2=25`

`+)4x-3=5`

`<=>4x=8<=>x=2`

`+)4x-3=-5`

`<=>4x=-2`

`<=>x=-1/2`

`b)x^2+10x+9=0`

`<=>x^2+x+9x+9=0`

`<=>x(x+1)+9(x+1)=0`

`<=>(x+1)(x+9)=0`

`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2+2x-6x-12=0`

`<=>x(x+2)-6(x+2)=0`

`<=>(x+2)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2+x-6x-6=0`

`<=>x(x+1)-6(x+1)=0`

`<=>(x+1)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

`e)4x^2-3x-1=0`

`<=>4x^2-4x+x-1=0`

`<=>4x(x-1)+(x-1)=0`

`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\) 

`f)x^4+4x^2-5=0`

`<=>x^4-x^2+5x^2-5=0`

`<=>x^2(x^2-1)+5(x^2-1)=0`

`<=>(x^2-1)(x^2+5)=0`

Vì `x^2+5>=5>0`

`=>x^2-1=0<=>x^2=1`

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

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

(2x + 5)(4x² - 9) = 0

<=> \(\left[{}\begin{matrix}2x+5=0\\4x^2-9=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x^2=\dfrac{9}{4}< =>x=\pm\dfrac{3}{2}\end{matrix}\right.\)

KL: \(x\in\left\{\dfrac{-5}{2};\pm\dfrac{3}{2}\right\}\)

    (4x^2-9)(x+2)=0

⇔ (4x^2-9)=0        ->   (2x-3)(2x+3)=0     ->  2x-3=0     -> x=3/2  

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

     (x+2)=0             ->    x=-2

Vậy tập nghiệm S={ 3/2;-3/2;-2}

(4x²-9)(x+2) = 0

<=> [(2x)² - 3²] (x+2) = 0

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

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

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

<=> 2x = -3 hoặc 2x = 3 hoặc x = -2

<=> 2x:2 = -3:2 hoặc 2x:2 = 3:2 hoặc x = -2

<=> x = -3/2 hoặc x = 3/2 hoặc x = -2

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik

a: (x^2+9)(9x^2-1)=0

=>9x^2-1=0

=>x^2=1/9

=>x=1/3 hoặc x=-1/3

b: (4x^2-9)(2^(x-1)-1)=0

=>4x^2-9=0 hoặc 2^(x-1)-1=0

=>x^2=9/4 hoặc x-1=0

=>x=1;x=3/2;x=-3/2

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

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

=>\(\left[{}\begin{matrix}3x+2=0\\3-x=0\\3+x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};3;-3\right\}\)

d: (3x+3)^2(4x-4^2)=0

=>3x+3=0 hoặc 4x-16=0

=>x=4 hoặc x=-1

e: \(2^{\left(x-5\right)\left(x+2\right)}=1\)

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

=>x-5=0 hoặc x+2=0

=>x=5 hoặc x=-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

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

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

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

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

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