Question

Viết thành bình phương của một tổng hay hiệu 

a) \(x^2-6x+9\)

b) \(25+10x+x^2\)

c) \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)

Answer 1

a ) \(x^2-6x+9\)

\(=x^2-2.x.3+3^2\)

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

b ) \(25+10x+x^2\)

\(5^2+2.5.x+x^2\)

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

c ) \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)

\(=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}.y^4+\left(y^4\right)^2\)

\(=\left(\frac{1}{3}-x^4\right)^2\)

Answer 2

a) \(x^2-6x+9=\left(x-3\right)^2\)

b) \(25+10x+x^2=\left(5+x\right)^2\)

c) \(\frac{1}{9}-\frac{1}{3}y+y^8=\left(\frac{1}{3}-y^4\right)^2\)