2-√5x + / sqrt{2x-1}=0

как решить?

1) x²-x+1≥0⇒x-любое, т. к. D<0

2x²-1≥0⇒x≤-1/√2 U x≥1/√2

x∈ (-∞; -1/√2] U [1/√2; ∞)

x²-x+1=2x²-1

x²+x-2=0

x1+x2=-1 U x1*x2=-2

x1=-2 U x2=1

2) x+2≥0⇒x≥-2

x-6≥0⇒x≥6

x∈[6; ∞)

x+2=4+2√ (x-6) + x-6

2√ (x-6) = 4

√ (x-6) = 2

x-6=4

x=6+4=10

3) x+2≥0⇒x≥-2⇒x∈[-2; ∞)

2x²+8x+7=x²+4x+4

x²+4x+3=0

x1+x2=-4 U x1*x2=3

x1=-3 не удов усл

x2=-1

4) x∈[0; ∞)

√x=a

a²-a-6=0

a1+a2=1 U a1*a2=-6

a1=-2⇒√x=-2 нет решения

a2=3⇒√x=3⇒x=9

5) x+2≥2⇒x≥-2⇒x∈[-2; ∞)

x²+3x=0⇒x (x+3) = 0

x=0

x=-3-не удов усл

х+2=0

х=-2

6) x²-2x-15-9=0

x²-2x-24=0

x1+x2=2 U x1*x2=-24

x1=-4 U x2=6

7) (х-5) (х+3) - 3 (х-5) = 0

(х-5) * (x+3-3) = 0

x (x-5) = 0

x=0

x-5=0⇒x=5

8) x²+6x+4=0

D=36-16=20

x1 = (-6-2√5) / 2=3-√5

x2=3+√5