Question

*+(5a-2ab)=3a²+2ab-b²​

Answer 1

Ответ:

5a-(a+b)

Объяснение:

дұрыс деп ойлайм

Answer 2

To solve the given equation, let's start by simplifying both sides of the equation:

Left side: *+(5a-2ab) = 5a - 2ab

Right side: 3a² + 2ab - b²

Now we can set these two expressions equal to each other and solve for the variable 'a':

5a - 2ab = 3a² + 2ab - b²

Rearranging the equation:

3a² + 2ab - b² - 5a + 2ab = 0

Combining like terms:

3a² + 4ab - b² - 5a = 0

Now we have a quadratic equation. Let's try to factorize it. To do this, we look for two terms that multiply to give the product of the coefficient of a² (3) and the constant term (-b²) and add up to give the coefficient of 'a' (-5).

The factors of 3 are 1 and 3, and the factors of -b² are -1 and b (or b and -b).

We can rewrite the middle term (4ab) as (-ab + 5ab):

3a² - ab + 5ab - b² = 0

Now we can group the terms:

(a² - ab) + (5ab - b²) = 0

Taking out the common factors:

a(a - b) + b(5a - b) = 0

Now we have a product of two terms that equals zero. According to the zero product property, if a product of two factors equals zero, then at least one of the factors must be zero.

Setting each factor equal to zero:

a - b = 0 --> a = b 5a - b = 0 --> 5a = b

We can substitute 'a' with 'b' in the second equation:

5b = b

Since 'b' can be any non-zero value, we can divide both sides of the equation by 'b' (assuming b ≠ 0):

5 = 1

However, this is a contradiction. Therefore, there is no solution that satisfies the given equation *+(5a-2ab) = 3a² + 2ab - b² for all values of 'a' and 'b'.