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= ( x2 - 4y2 ) - ( 2x + 4y )
= ( x - 2y ) ( x + 2y ) - 2 ( x - 2y )
= ( x - 2y ) ( x + 2y - 2 )
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
a) x^2 - 5xy +4y^2= x^2 -xy -4xy+4y^2= (x^2-xy) - (4xy - 4y^2)= x(x-y)-4y(x-y)=(x-y)*(x - 4y)
b) x^2 -y^4+9y -x(9+y-y^3= x^2-y^4 +9y-9x-xy+xy^3= (x^2-xy)-(9x-9y)+(xy^3-y^4)=x(x-y)-9(x-y)+y^3(x-y)=(x-y)*(y^3+x-9)
d) 2u^2+2v^2-5uv=(2u^2-4uv)+(2v^2-uv)=2u(u-2v)+v(2v-u)= 2u(u-2v)-v(u-2v)=(u-2v)*(2u-v)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
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)
a)
\(4x^2-9y^2+6x-9y=\left(2x-3y\right)\left(2x+3\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b)
\(1-2x+2yz+x^2-y^2-z^2=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\) (đổi dấu)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
c)
\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5\left(x+1\right)+3\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)
\(a,=\left(x-2y\right)\left(x+2y\right)-2\left(x-2y\right)=\left(x-2y\right)\left(x+2y-2\right)\\ b,=\left(x^2+3y\right)^2-1=\left(x^2+3y-1\right)\left(x^2+3y+1\right)\)
\(b)x^4+6x^2y+9y^2-1\\=(x^4+6x^2y+9y^2)-1\\=(x^2+3y)^2-1\\=(x^2+3y-1)(x^2+3y+1)\\a) x^2-2x+4y^2+4y\\=(x^2-4y^2)-(2x-4y)\\=(x-2y)(x+2y)-2(x-2y)\\=(x-2y)(x+2y-2)\)
\(x^2-2x-4y^2+4y=x^2-2x+1-4y^2+4y-1=\left(x-1\right)^2-\left(2y-1\right)^2=\left(x-1-2y+1\right)\left(x-1+2y-1\right)=\left(x-2y\right)\left(x+2y-2\right)\)
\(x^4+6x^2y+9y^2-1=\left(x^2+3y\right)^2-1=\left(x^2+3y-1\right)\left(x^2+3y+1\right)\)