18 Апр 2021 в 19:53
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Ответы
1

To solve the first equation x+2-6(x-6)=6(6-x)+6:

Distribute the -6 to both terms inside the parentheses: x + 2 - 6x + 36 = 36 - 6x + 6Combine like terms on both sides: x - 6x + 38 = 36 - 6x + 6Simplify further: -5x + 38 = -6x + 42Add 6x to both sides to get the variable terms on one side: x + 38 = 42Subtract 38 from both sides: x = 4
Therefore, the solution to the equation x+2-6(x-6)=6(6-x)+6 is x = 4.

For the second equation 2x^2 - 5x - 7 = 0:

This is a quadratic equation, which can be factored or solved using the quadratic formula.The factorization may not be immediately obvious, so let's use the quadratic formula: x = (-(-5) ± sqrt((-5)^2 - 4 2 -7)) / 2 * 2Simplify within the square root: x = (5 ± sqrt(25 + 56)) / 4Further simplification: x = (5 ± sqrt(81)) / 4Complete the square root: x = (5 ± 9) / 4
Now solve for x with both possibilities:First, x = (5 + 9) / 4 = 14 / 4 = 3.5Second, x = (5 - 9) / 4 = -4 / 4 = -1
Therefore, the solutions to the equation 2x^2 - 5x - 7 = 0 are x = 3.5 and x = -1.
17 Апр 2024 в 18:56
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