Question

Bai 1; Viet thanh tong hoac hieu cua 2 binh phuong:

a, x²+10x+26+y²+2y

b, x²-2xy+2y²+2y+1

c,z²-6z+5-t²-4t

d, 4x²-12x-y²+2y+1 (Giup minh vs nha, thank nhiu)

Answer 1

a) \(x^2+10x+26+y^2+2y\)

= \(x^2+10x+25+y^2+2y+1\)

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

b) \(x^2-2xy+2y^2+2y+1\)

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

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

c) \(z^2-6z+5-t^2-4t\)

= \(z^2-6z+9-\left(t^2+4t+4\right)\)

= \(\left(z-3\right)^2-\left(t+2\right)^2\)

d) \(4x^2-12x-y^2+2y+1\)

Hình như câu này sai đề -_-

Answer 2

a, \(x^2+10x+26+y^2+2y\)

\(=\left(x^2+2.x.5+5^2\right)+\left(1^2+2.1.y+y^2\right)\)

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

b, \(x^2-2xy+2y^2+2y+1\)

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

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

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

c,\(z^2 -6z+5-t^2-4t\)

\(=-\left(t^2+4t-z^2+6z-5\right)\)

\(=-\left(t^2+2.t.2+2^2-z^2+2.z.3-3^2\right)\)

\(=-\left(\left(t^2+2.t.2+2^2\right)-\left(z^2-2.z.3+3^2\right)\right)\)

\(=-\left(\left(t+2\right)^2-\left(z-3\right)^2\right)\)

\(=\left(z-3\right)^2-\left(t+2\right)^2\)

d, Không biết làm hihi :)