1) \(\Leftrightarrow\left(x-4\right)\left(x+4\right)-x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4-x\right)=0\)

\(\Leftrightarrow\left(x-4\right)4=0\)

\(\Leftrightarrow x=4\)

2) \(\left(x+3\right)^2-\left(x-3\right)\left(x+5\right)=x^2+6x+9-x^2-2x+15=4x+24\)

3) \(2x^3+3x^2-2x+a=2x^2\left(x-2\right)+7x\left(x-2\right)+16\left(x-2\right)+32+a\)

Để \(2x^3+3x^2-2x+a⋮x-2\) thì \(32+a=0\Leftrightarrow a=-32\)

1. 

x² - 16 - x(x - 4) = 0

<=> (x² - 4²) - x(x - 4) = 0

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

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

<=> 4(x + 4) = 0

<=> x + 4 = 0

<=> x = -4

2.

(x + 3)² - (x - 3)(x + 5)

= x² + 6x + 9 - (x² + 5x - 3x - 15)

= x² + 6x + 9 - x² + 5x - 3x - 15

= x² - x² + 6x + 5x - 3x + 9 - 15

= 8x - 6

⇒x=4

⇒⇒ 2x = -4 ⇒ x = - 2

Vậy  hoặc .

3. Ta có : 2x³ + 3x² - 2x + a = (x - 2)(2x² + 7x + 12) + (a - 24)
Để phép chia trên là phép chia hết thì a - 24 = 0 => a = 24

Còn bài 2 mình khum biéc làm 😢😢😢

Bài 1: 

b: \(3x-6=x^2-16\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

b: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)

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

a: 3x^3+2x^2-7x+a chia hêt cho 3x-1

=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1

=>a-2=0

=>a=2

c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4

=>3a+19=4

=>3a=-15

=>a=-5

d: 2x^3-x^2+ax+b chiahêt cho x^2-1

=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1

=>a+2=0 và b-1=0

=>a=-2 và b=1

1.

\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+4}{x-3}\)

b.

\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)

\(\Rightarrow x=10\) (thỏa mãn)

2.

\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)