Задача Задача №2405927
А
А
Let's simplify each expression step by step.
a) \( 6x(x - 3) - x(2 - x) \)
- Distribute the terms: \[ 6x(x) - 6x(3) - x(2) + x(x) = 6x^2 - 18x - 2x + x^2 \]
- Combine like terms: \[ 6x^2 + x^2 - 18x - 2x = 7x^2 - 20x \]
So, the simplified form is: \[ \boxed{7x^2 - 20x} \]
b) \( -a(3a - 5) + 4a(a^2 - a) \)
- Distribute the terms: \[ -a(3a) + a(5) + 4a(a^2) - 4a(a) = -3a^2 + 5a + 4a^3 - 4a^2 \]
- Combine like terms: \[ 4a^3 - 3a^2 - 4a^2 + 5a = 4a^3 - 7a^2 + 5a \]
So, the simplified form is: \[ \boxed{4a^3 - 7a^2 + 5a} \]
c) \( ax(2x - 3a) - x(ax + 5a^3) \)
- Distribute the terms: \[ ax(2x) - ax(3a) - x(ax) - x(5a^3) = 2ax^2 - 3a^2x - ax^2 - 5a^3x \]
- Combine like terms: \[ 2ax^2 - ax^2 - 3a^2x - 5a^3x = (2a - a)x^2 - (3a^2 + 5a^3)x = ax^2 - (3a^2 + 5a^3)x \]
So, the simplified form is: \[ \boxed{ax^2 - (3a^2 + 5a^3)x} \]
d) \( -4m'(n^2 - m') + 3n'(m^0 - n^0) \)
- Recall that \( m^0 = 1 \) and \( n^0 = 1 \), thus: \[ -4m'(n^2 - m') + 3n'(1 - 1) = -4m'(n^2 - m') + 3n'(0) \] This simplifies to: \[ -4m'(n^2 - m') + 0 = -4m'(n^2 - m') \]
So, the simplified form is: \[ \boxed{-4m'(n^2 - m')} \]
These are the simplified forms of the given expressions.
