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8mixa1111
@8mixa1111
July 2022
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2-√5x+ \sqrt{2x-1}=0
как решить ?????????????????
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Answers & Comments
LAGGGER231
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
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Answers & Comments
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