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
1.
x2−16+x(x−4)=0
(x2−16)+x(x−4)=0
(x+4)(x−4)+x(x−4)=0
(x−4)(x+4+x)=0
(x−4)(2x+4)=0
⇒x−4=0⇒x=4
⇒2x + 4=0 ⇒ 2x = -4 ⇒ x = - 2
Vậy x=−2 hoặc x=4.
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 😢😢😢