a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]:\left(x+2\right)\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

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

help

D

a) x(4x²-1)=0

=>x(2x-1)(2x+1)=0

=>\(\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

vậy x\(\in\) {\(\dfrac{-1}{2}\) ;0;\(\dfrac{1}{2}\) }

c)x³-x²-x+1=0

=>(x³-x²)-(x-1)=0

=>x²(x-1)-(x-1)=0

=>(x-1)(x²-1)=0

=>\(\left[{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)

Bổ sung thêm \(x^2=1\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\).

8: \(x^3-4x^2+8x-32=0\)

=>\(x^2\left(x-4\right)+8\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2+8\right)=0\)

=>x-4=0

=>x=4

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

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

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

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

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

=>x∈{0;2;-3}

10: \(x^2-10x+16=0\)

=>\(x^2-2x-8x+16=0\)

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

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

=>x=2 hoặc x=8

11: \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

=>\(2x^2+3\left(x^2-1\right)=5x^2+5x\)

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

=>5x=-3

=>\(x=-\frac35\)

12: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=0\)

=>\(x^2-9-\left(x^2+3x-10\right)=0\)

=>\(x^2-9-x^2-3x+10=0\)

=>-3x+1=0

=>-3x=-1

=>\(x=\frac13\)

Q(x)=2P(x)

=>2x^3-4x^2+5x=2x^3-4x^2+6

=>5x=6

=>x=6/5

a) Đa thức thương  x 2  – 6x + 9.

b) Đa thức thương 2 x 2  – 5.

c) Đa thức thương  x 2  + 4x + 3 và đa thức dư -12.

d) Đa thức x + 5 và đa thức dư x – 4.

1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)