Question

x^2-3|x|-5=10-5|x| ;
|x^2+5x|-x^2=(x+2)(2-x)-5 ;
|x^2-2x+4|=2|x|​

Answer 1

a^2 - b^2 = (a - b)(a + b)

|x|^2 = x^2

|x| >= 0

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X^2-3|x|-5=10-5|x|

|x|^2 = x^2

|x|^2 + 2|x| - 15 = 0

D = 4 + 60 = 64

|x₁₂| = (-2 +- 8)/2 = -5   3

|x| = -5 нет |x|>=0

|x| = 3

x1 = 3

x2 = -3

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|x^2+5x|-x^2=(x+2)(2-x)-5

|x^2+5x|-x^2=4 - x^2 -5

|x^2+5x| = -1

Решений нет модуль неотрицателен

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|x^2-2x+4|=2|x|​

|a| = |b| ⇔ a^2 = b^2

(x^2-2x+4)^2=(2x)^2

(x^2-2x+4)^2 - (2x)^2 = 0

​(x^2-2x+4-2x)(x^2-2x+4+2x) = 0

(x^2 - 4x + 4)(x^2 + 4) = 0

x^2 + 4 > 0

(x - 2)^2 = 0

x = 2