\(x-1=\sqrt[3]{9}+\sqrt[3]{3}\)

=> \(\left(x-1\right)^3=9+3+3.3.\left(x-1\right)\)

<=> \(x^3-3x^2+3x+1=3+9x\)

<=> \(x^3-3x^2-6x-3=-1\)

=> \(P=-1+2020=2019\)

\(a,\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\\ b,\Rightarrow x\left(x^2-13\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\\ c,\Rightarrow3x\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\Rightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\\ e,\Rightarrow\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(a,=\left(x-2\right)^2\\ b,=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\\ c,=\left(1-2x\right)\left(1+2x+4x^2\right)\\ d,=\left(x+1\right)^3\\ e,Sửa:\left(x+y\right)^2-9x^2=\left(x+y-3x\right)\left(x+y+3x\right)\\ =\left(y-2x\right)\left(4x+y\right)\\ f,=\left(x+3\right)^2\\ g,=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\\ h,=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)

a) \(\left(x-2\right)^2\)

b) \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

c) \(\left(1-2x\right)\left(1+2x+4x^2\right)\)

d) \(\left(x+1\right)^3\)

e) \(\left(x+y-3\sqrt{x}\right)\left(x+y+3\sqrt{x}\right)\)

f) \(\left(x+3\right)^2\)

g) \(-\left(x-5\right)^2\)

h) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

gấp

Bạn cần làm gì với đa thức này?

6: \(\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)\)

\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}\)

\(=\frac{\left(2x-5\right)\left(x^2+3\right)}{2x-5}=x^2+3\)

2: \(\frac{2x^4-5x^2+x^3-3-3x}{x^2-3}\)

\(=\frac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)

\(=\frac{2x^2\left(x^2-3\right)+x\cdot\left(x^2-3\right)+\left(x^2-3\right)}{x^2-3}=2x^2+x+1\)

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

\(=\frac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}=\frac{\left(2x^2-x+1\right)\left(x+3\right)}{2x^2-x+1}\)

=x+3

3: \(\left(x-y-z\right)^5:\left(x-y-z\right)^3=\left(x-y-z\right)^{5-3}=\left(x-y-z\right)^2\)

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

\(=\frac{x^2\left(x-3\right)+\left(x-3\right)}{x-3}=x^2+1\)

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...

sao lại trả lời lại nhỉ ??