Question

1)х³(х²-у²)-х³(х²+у²)+2х³у²-72) (3ху-5)²-(5ху-3)²+16(х²у²-2)-1​

Answer 1

Объяснение:

1)

$x^3*(x^2-y^2)-x^3*(x^2+y^2)+2x^3y^2-7=x^3*(x^2-y^2-x^2-y^2+2y^2)-7=\\x^3*0-7=0-7=-7.$

2)

$(3xy-5)^2-(5xy-3)^2+16*(x^2y^2-2)-1=\\=9x^2y^2-30xy+25-(25x^2y^2-30xy+9)+16x^2y^2-32-1=\\=9x^2y^2-30xy+25-25x^2y^2+30xy-9+16x^2y^2-32-1=-17.$

Answer 2

Let's simplify both of the given expressions:

1. $x^3(x^2 - y^2) - x^3(x^2 + y^2) + 2x^3y^2 - 7$

First, factor out $x^3$ from the first two terms:

$x^3(x^2 - y^2 - x^2 - y^2) + 2x^3y^2 - 7$

Now, simplify the terms inside the parentheses:

$x^3(-2y^2) + 2x^3y^2 - 7$

Now, distribute $x^3$ into the remaining terms:

$-2x^3y^2 + 2x^3y^2 - 7$

Now, the $2x^3y^2$ terms cancel each other out:

$-7$

So, the simplified expression is just $-7$.

2. $(3xy - 5)^2 - (5xy - 3)^2 + 16(x^2y^2 - 2) - 1$

Let's simplify this expression step by step:

Expand the squares:

$(9x^2y^2 - 30xy + 25) - (25x^2y^2 - 30xy + 9) + 16x^2y^2 - 32 - 1$

Now, subtract the second expression from the first:

$(9x^2y^2 - 30xy + 25) - (25x^2y^2 - 30xy + 9)$

Distribute the negative sign inside the parentheses:

$9x^2y^2 - 30xy + 25 - 25x^2y^2 + 30xy - 9$

Now, combine like terms:

$(9x^2y^2 - 25x^2y^2) + (-30xy + 30xy) + (25 - 9)$

Simplify each part separately:

$-16x^2y^2 + 0 + 16$

So, the simplified expression is $16 - 16x^2y^2$.