Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)

\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)

\(=\left(3x-5y\right)\left(2x-y\right)\)

b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)

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

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

\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)

\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)

a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)

e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)

\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)

\(a,A=x^2+7x+7y-y^2\\ =x^2-y^2+7x+7y\\ =\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\\ =\left(x+y\right)\left(x-y+7\right)\)

\(b,B=x^2+2xy+y^2-3x-3y\\ =\left(x+y\right)^2-3\left(x+y\right)\\ =\left(x+y\right)\left(x+y-3\right)\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

2:

a: \(3xy^2-3x^3-6xy+3x\)

\(=3x\cdot\left(y^2-2y+1-x^2\right)\)

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

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

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

\(=x\left(16+4xy-4y^2-x^2\right)\)

\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

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

d: \(xz-x^2-yz+2xy-y^2\)

=z(x-y)-\(\left(x^2-2xy+y^2\right)\)

=\(z\left(x-y\right)-\left(x-y\right)^2\)

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

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

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

f: \(x^4-x^3-10x^2+2x+4\)

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

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

g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)

\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)

\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)

=(x-1)(x^2+1)(11-5y-1)(11+5y+1)

=(x-1)(x^2+1)(10-5y)(12+5y)

=5(2-y)(x-1)(x^2+1)(5y+12)

`9-x^2-2xy-y^2`

`=9-(x^2+2xy+y^2)`

`=3^2-(x+y)^2`

`=(3+x+y)(3-x-y)`

9-x²-2xy-y²=9-(x²+2xy+y²)=3²-(x+y)²=(3-x-y)(3+x+y)

a, 7x - 14

= 7(x-2)

b, 2x - 2y + \(x^2\)- xy 

= (2x-2y) + (\(x^2\)-xy)

= 2(x-y) + x(x-y)

= (x-y)(2+x)

c, 6x + 12

= 6(x+2)

\(a,=7\left(x-2\right)\\ b,=2\left(x-y\right)+x\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=6\left(x+2\right)\\ d,\text{Sai đề}\)

\(a,Sửa:x^2+4xy-9+4y^2=\left(x+2y\right)^2-9=\left(x+2y-3\right)\left(x+2y+3\right)\\ b,=\left(x-6y\right)^2-1=\left(x-6y-1\right)\left(x-6y+1\right)\\ c,=36-\left(x-5y\right)^2=\left(6-x+5y\right)\left(6+x-5y\right)\)

\(A=x^2-y^2+7x+7y\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+7\right)\)

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

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

\(=x\left(2x-1\right)^2\)

\(C=x^2-6xy+9y^2-9\)

\(=\left(x-3y\right)^2-9\)

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

A=\(x^2+7x+7y-y^2=\left(x^2-y^2\right)+\left(7x+7y\right)=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)=\left(x+y\right)\left(x-y+7\right)\)

B=\(4x^3-4x^2+x=x\left(4x^2-4x+1\right)=x\left(2x-1\right)^2\)

C=\(x^2+9y^2-9-6xy=\left(x^2-6xy+9y^2\right)-9=\left(x-3y\right)^2-3^2=\left(x-3y-3\right)\left(x-3y+3\right)\)