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

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

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

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

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

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

b: \(x^2-2xy+y^2-z^2\)

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

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

d: \(x^2+4x+3=\left(x+3\right)\left(x+1\right)\)

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

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

g) \(6x^2+7x-5\)

=\(6x^2+10x-3x-5\)

=\(\left(6x^2+10x\right)-\left(3x+5\right)\)

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

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

1D  2C

Câu 1: D

Câu 2: C

A

a)2x^2+xy-y^2-x+2y-1

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

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

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

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

\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

a,=x(x−2)b,=2b(x−3y)+a(x−3y)=(a+2b)(x−3y)c,=x(x2+2xy+y2−4)=x[(x+y)2−4]=x(x+y+2)(x+y−2)d,=4−(x+y)2=(2−x−y)(2+x+y)đ,=5(x−y)(x+y)+3(x+y)2=(x+y)(5x−5y+3x+3y)=(x+y)(8x−2y)=2(4x−y)(x+y)e,=3x(2xy−3)b,=x(4x2−4xy+y2−4)=x[(2x−y)2−4]=x(2x−y−2)(2x−y+2)f,=(x+y)2−z2=(x+y−z)(x+y+z)