Question

Answer 1

Bài 22:
1.

PT $\Leftrightarrow \frac{3a^3(x^2-1)^4}{3a^3(x^2-1)^3}=15$
$\Leftrightarrow x^2-1=15$

$\Leftrightarrow x^2=16$

$\Leftrightarrow x=\pm 4$ (tm)

2.

PT $\Leftrightarrow (2x-1)^{m+2-(m-1)}-3^3=0$

$\Leftrightarrow (2x-1)^3=3^3$

$\Leftrightarrow 2x-1=3$

$\Leftrightarrow x=2$

3.

PT $\Leftrightarrow (3x^2-4)-(3x+1)-3x^2=0$

$\Leftrightarrow -4-(3x+1)=0$

$\Leftrightarrow 3x+1=-4$

$\Leftrightarrow x=\frac{-5}{3}$

Answer 2

 

Bài 23:

1.

\(3x^5-x^4-2x^3+x^2+4x+5=3x^3(x^2-2x+2)+5x^4-8x^3+x^2+4x+5\)

\(=3x^3(x^2-2x+2)+5x^2(x^2-2x+2)+2x^3-9x^2+4x+5\)

\(=(3x^3+5x^2)(x^2-2x+2)+2x(x^2-2x+2)-5x^2+5\)

\(=(x^2-2x+2)(3x^3+5x^2+2x)-5(x^2-2x+2)-10x+15\)

$=(x^2-2x+2)(3x^3+5x^2+2x-5)-10x+15$

Vậy đa thức dư là $-10x+15=0\Leftrihgtarrow x=\frac{3}{2}$

2.

\(x^5+2x^4+3x^2+x-3=x^3(x^2+1)+x^2(x^2+1)+2(x^2+1)+x-5\)

$=(x^2+1)(x^3+x^2+2)+x-5$

Đa thức dư là $x-5=0\Leftrightarrow x=5$