\(B=x^2-x^3+x^3+27=x^2+27\)

\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)

Ui em k rõ là giải phương trình hay rút gọn nữa :((

a) \(6x^2-15x\)

b) \(x^2+5x+4\)

c) \(49-x^2\)

d) \(x^2+4x+4\)

e) \(9-12x+4x^2\)

f) \(x^3-8\)

\(a,=6x^2-15x\\ b,=x^2+5x+4\\ c,=49-x^2\\ d,=x^2+4x+4\\ e,=9-12x+4x^2\\ f,=x^3-8\)

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

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

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)