Question

Viết các đa thức sau dưới dạng bình phương của 1 tổng hoặc 1 hiệu

a)x²+ 4x + 4

b) 9x² + 42x + 49

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

Answer 1

a)\(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)

b)\(9x^2+42x+49=\left(3x\right)^2+2\cdot3x\cdot7+7^2=\left(3x+7\right)^2\)

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

Answer 2

a) \(x^2+2.2x+2^2\)

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

b)\(\left(3x\right)^2+2.3.7x+7^2\)

\(=\left(3x+7\right)^2\)

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

\(=\left(\dfrac{1}{3}-y^4\right)^2\)