1)x²-3x+2|x-2|=0 2)sinα-4cosα/2sinα+cosα если tg α/2 = 2 3(log₂3+log₃16+4)(log₂3-2log₁₂3)log₃2-log₂3 4)y=1/корень(x²-4) 5)1+cosα+cos2α+cos3α/sin2α+2sinαcos2α=ctgα

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вопрос

1) (x-2)(x-1) = 0
x = 2 or x = 1

2) Replace sinα and cosα in the given expression:
sinα - 4cosα / 2sinα + cosα = sinα - 4√(1 - sin²α) / 2sinα + cosα
= sinα - 4√(1 - sin²α) / 2sinα + √(1 - sin²α)

Given that tg(α/2) = 2, we can find sinα and cosα:
sinα = 2/(1+2²) = 2/5
cosα = √(1 - (2/5)²) = √(21)/5

Substitute sinα and cosα back into the expression:
(2/5) - 4(√(21)/5) / 2(2/5) + √(21)/5
= (2/5) - 4(√21)/5 / (4/5) + √21/5
= (2 - 4√21) / 4 + √21
= (2 - 4√21) / 4 + √21

3) Simplify the expression:
3(log₂3 + log₃16 + 4)(log₂3 - 2log₁₂3)log₃2 - log₂3
= 3(log₂3 + 4 + 2log₃2)(log₂3 - log₂3²) - log₂3
= 3(log₂3 + 4 + 2)(log₂3 - 3) - log₂3
= 3(4 + log₂3 + 8)(log₂3 - 3) - log₂3
= 3(12 + log₂3)(log₂3 - 3) - log₂3

4) y = 1/√(x² - 4)
This is the equation of a hyperbola:

5) 1 + cosα + cos2α + cos3α / sin2α + 2sinαcos2α = ctg(α)
To simplify the expression, use trigonometric identities and properties of cotangent.