Question

(5y-1)²+(3+5y)(3-5) розкрити дужки

Answer 1

$\displaystyle\bf\\(5y-1)^{2} +(3+5y)(3-5y)=\\\\=(5y)^{2}-2\cdot 5y\cdot 1+1^{2}+3^{2} -(5y)^{2} =\\\\=25y^{2}-10y+1+9-25y^{2} =\\\\=\underbrace{(25y^{2} -25y^{2})}_{0}-10y+\underbrace{(1+9)}_{10} =10-10y$

Answer 2

Ответ:

10 - 10у

Объяснение:

(5у - 1)² + (3 + 5у)(3 - 5у) =

(25у² - 10у + 1) + (9 - 15у + 15у - 25у²) =

25у² - 10у + 1 + 9 - 25у² = 10 - 10у

Answer 3

To expand the given expression $(5y-1)^2 + (3+5y)(3-5)$, you can use the formula for squaring a binomial $(a-b)^2 = a^2 - 2ab + b^2$. Let's apply this formula to the first term:

$(5y-1)^2 = (5y)^2 - 2(5y)(1) + (1)^2$

Simplifying each term, we get:

$(5y-1)^2 = 25y^2 - 10y + 1$

Now, let's simplify the second term $(3+5y)(3-5)$:

$(3+5y)(3-5)$

Using the distributive property (FOIL method), we multiply the terms:

$= 3 \cdot 3 + 3 \cdot (-5) + 5y \cdot 3 + 5y \cdot (-5)$

Simplifying each term, we get:

$= 9 - 15 + 15y - 25y$

Combine like terms:

$= -6 - 10y$

Now, put these two simplified expressions back together:

$(5y-1)^2 + (3+5y)(3-5) = 25y^2 - 10y + 1 - 6 - 10y$

Combine like terms:

$= 25y^2 - 20y - 5$

So, $(5y-1)^2 + (3+5y)(3-5)$ simplifies to $25y^2 - 20y - 5$.