Question

viết các biểu thức dưới dạng bình phương của 1 tổng , hiệu

4x²y⁴ - 4xy³ + y²

(x - 2y) - 4(x - 2y )+4

25x² - 5xy +\(\dfrac{1}{4}\) y²

Answer 1

4x²y⁴ - 4xy³ + y²

= (2xy²)² - 2.2xy².y + y²

= (2xy² - y)²

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Sửa đề:

(x - 2y)² - 4(x - 2y) + 4

= (x - 2y)² - 2.(x - 2y).2 + 2²

= (x - 2y - 2)²

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25x² - 5xy + 1/4 y²

= (5x)² - 2.5xy.y/2 + (y/2)²

= (5x - y/2)²

Answer 2

\(4x^2y^4-4xy^3+y^2\)

\(=\left(2xy^2\right)^2-2\cdot2xy^2\cdot y+y^2\)

\(=\left(2xy^2-y\right)^2\)

_____
\(\left(x-2y\right)^2-4\left(x-2y\right)+4\)

\(=\left(x-2y\right)^2-2\cdot\left(x-2y\right)\cdot2+2^2\)

\(=\left[\left(x-2y\right)-2\right]^2\)

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

____

\(25x^2-5xy+\dfrac{1}{4}y^2\)

\(=\left(5x\right)^2-2\cdot\dfrac{5}{2}xy+\left(\dfrac{1}{2}y\right)^2\)

\(=\left(5x\right)^2-2\cdot\dfrac{1}{2}y\cdot5x+\left(\dfrac{1}{2}y\right)^2\)

\(=\left(5x-\dfrac{1}{2}y\right)^2\)