1) \(3x^2-9=3\left(x^2-3\right)\)

2) \(\dfrac{1}{2}x^2-2y^2\)

\(=\dfrac{1}{2}\left(x^2-4y^2\right)\)

\(=\dfrac{1}{2}\left(x-2y\right)\left(x +2y\right)\)

3) \(3x^2-12y^2\)

\(=3\left(x^2-4y^2\right)\)

\(=3\left(x-2y\right)\left(x+2y\right)\)

4) \(\dfrac{1}{3}x^2y^2-3x^2\)

\(=\dfrac{1}{3}x^2\left(y^2-9\right)\)

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

#Toru

\(a,x^4+5x^3-8x-40=x^3\left(x+5\right)-8\left(x+5\right)\\ =\left(x^3-8\right)\left(x+5\right)=\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)\\ b,3x^2-6x-12y^2+3=3\left(x^2-2x-4y^2+1\right)\\ =3\left[\left(x-1\right)^2-4y^2\right]=3\left(x-2y-1\right)\left(x+2y-1\right)\)

\(ĐK:x\ge-2;y\le4\)

\(PT\left(1\right)\Leftrightarrow\left(x^3-3x^2+3x-1\right)-\left(y^3-6y^2+12y-8\right)=0\\ \Leftrightarrow\left(x-1\right)^3-\left(y-2\right)^3=0\\ \Leftrightarrow\left(x-y+1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(y-2\right)+\left(y-2\right)^2\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x-y+1=0\\x^2-4x+xy+y^2-5y+7=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow\left(x^2+\dfrac{1}{4}y^2+4+xy-2y-4x\right)+\dfrac{3}{4}y^2-3y+3=0\\ \Leftrightarrow\left(x+\dfrac{1}{2}y-2\right)^2+\dfrac{3}{4}\left(y^2-4y+4\right)=0\\ \Leftrightarrow\left(x+\dfrac{1}{2}y-2\right)^2+\dfrac{3}{4}\left(y-2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Thay \(x=1;y=2\) vào PT(2) ta thấy ko thỏa mãn

Với \(x-y+1=0\Leftrightarrow y=x+1\), thay vào PT(2)

\(\Leftrightarrow\sqrt{x+2}+\sqrt{3-x}=x^3+x^2-4x-1\left(-2\le x\le3\right)\\ \Leftrightarrow\sqrt{x+2}+\sqrt{3-x}-3=x^3+x^2-4x-4\\ \Leftrightarrow\dfrac{2\sqrt{\left(x+2\right)\left(3-x\right)}-4}{\sqrt{x+2}+\sqrt{3-x}+3}=\left(x+1\right)\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow\dfrac{2\left[\left(x+2\right)\left(3-x\right)-4\right]}{\left(\sqrt{x+2}+\sqrt{3-x}+3\right)\left(\sqrt{\left(x+2\right)\left(3-x\right)}+2\right)}=\left(x^2-x-2\right)\left(x+2\right)\\ \Leftrightarrow\left(x^2-x-2\right)\left(x+2\right)+\dfrac{2\left(x^2-x-2\right)}{\left(\sqrt{x+2}+\sqrt{3-x}+3\right)\left(\sqrt{\left(x+2\right)\left(3-x\right)}+2\right)}=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left[x+2+\dfrac{1}{\left(\sqrt{x+2}+\sqrt{3-x}+3\right)\left(\sqrt{\left(x+2\right)\left(3-x\right)}+2\right)}\right]=0\)

Với \(x\ge-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=0\\x=2\Rightarrow x=3\end{matrix}\right.\left(tm\right)\)

Vậy HPT có nghiệm \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(2;3\right)\right\}\)

a) 

P = (x² -4x )- (3x² -4+5x)

P = x² - 4x - 3x²  +4 - 5x

P= (x²  - 3x² ) + (-4x - 5x ) + 4

P= -2x²   - 9x + 4

b)

Q= (-12y⁵ + y⁴ -1 )+ ( 14y⁴ + 6y⁵ -3 )

Q= -12y⁵ + y⁴  - 1 + 14y⁴ + 6y⁵  - 3 

Q= ( -12y⁵ + 6y⁵ )+ ( y⁴ + 14y⁴ ) + (-1-3)

Q= -6y⁵  + 15y⁴ -4 

chúc bn hok tốt !~##

a: =(2x-3y)^2-4(2x-3y)

=(2x-3y)(2x-3y-4)

b: =3x^2+21x-x-7

=(x+7)(3x-1)

c: =(3x-1)^4+2(3x-1)^2+1

=[(3x-1)^2+1]^2

d: =2x^3-2x^2-x^2+x+x-1

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