Question

59.)1) a +(b-(c-d));2) a-(b-(c-d));3) a-((b-c)-d);4) a-(b+(C-(d - k))).​

Answer 1

Ответ:

$1)\ \ a+(b-(c-d))=a+(b-c+d)=a+b-c+d\\\\2)\ \ a-(b-(c-d))=a-(b-c+d)=a-b+c-d\\\\3)\ \ a-((b-c)-d)=a-(b-c-d)=a-b+c+d\\\\4)\ \ a-(b+(c-(d-k)))=a-(b+(c-d+k))=a-(b+c-d+k)=\\\\=a-b-c+d-k$

Answer 2

Let's break down the expressions:

1. a + (b - (c - d)): First, we subtract (c - d), then subtract the result from b, and finally add the outcome to a.

2. a - (b - (c - d)): Similar to the first expression, we subtract (c - d) from c, then subtract the result from b, and finally subtract the outcome from a.

3. a - ((b - c) - d): Here, we subtract c from b, then subtract the result from d, and finally subtract the outcome from a.

4. a - (b + (C - (d - k))): This expression is similar to the second one, but with an additional nested expression: (d - k). First, we subtract (d - k) from d, then subtract the result from C, and finally subtract the outcome from b. Finally, we subtract the entire result from a.

Remember that the uppercase letters "C" and "K" in expression 4) are different variables from the lowercase "c" and "k" used in the other expressions.