
A) 10a - (-(5a + 20)) = 5(3a + 4); б) 12b - (25 - (6b - 11)) = 18(b - 2);B) a2 + 7a + 10 = (a +
2)(a + 5);г) -(-7a) - (6 + 5a) = 2(a - 3);д) 36 - (--(9b - 15)) = 3(3b + 7);e) n2 - 9n + 20 = (n - 4)(n - 5).

Ответы на вопрос

Ответ:



Ответ:
а) 10а-(-(5а+20)) =5(3а+4)
10а+5а+20=15а+20
10а+5а-15а=20-20
0*х=0
Ответ: нет корней



Let's solve each of these equations step by step:
A) 10a - (-(5a + 20)) = 5(3a + 4)
First, distribute the negative sign inside the parentheses: 10a - (-5a - 20) = 5(3a + 4)
Now, eliminate the double negative: 10a + 5a + 20 = 5(3a + 4)
Combine like terms on the left side: 15a + 20 = 15a + 20
Now, subtract 15a from both sides to isolate the constant: 15a - 15a + 20 = 15a - 15a + 20
This simplifies to: 20 = 20
The equation is true for all values of "a."
B) 12b - (25 - (6b - 11)) = 18(b - 2)
First, simplify the innermost parentheses: 12b - (25 - 6b + 11) = 18(b - 2)
Now, simplify the inner parentheses: 12b - (36 - 6b) = 18(b - 2)
Distribute the negative sign inside the outer parentheses: 12b - 36 + 6b = 18(b - 2)
Combine like terms on both sides of the equation: 18b - 36 = 18b - 36
Now, subtract 18b from both sides to isolate the constant: 18b - 18b - 36 = 18b - 18b - 36
This simplifies to: -36 = -36
The equation is true for all values of "b."
C) a^2 + 7a + 10 = (a + 2)(a + 5)
This equation is already simplified, and it represents the factorization of the left side of the equation.
D) -(-7a) - (6 + 5a) = 2(a - 3)
First, simplify the double negative: 7a - (6 + 5a) = 2(a - 3)
Now, distribute the negative sign inside the parentheses: 7a - 6 - 5a = 2(a - 3)
Combine like terms on both sides of the equation: 2a - 6 = 2(a - 3)
Distribute 2 on the right side: 2a - 6 = 2a - 6
Now, add 6 to both sides to isolate the constant: 2a - 6 + 6 = 2a - 6 + 6
This simplifies to: 2a = 2a
The equation is true for all values of "a."
E) n^2 - 9n + 20 = (n - 4)(n - 5)
This equation is already simplified, and it represents the factorization of the left side of the equation.
In summary, all of the provided equations are identities, which means they are true for all values of the respective variables (a, b, and n).


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