Question

Rozwiąż równanie : x³-3√2x²+√2x-6=0; 1-x³=x²-x; x³-5x-4=0 wskazówka zapisz -5x jako -x-4x; x³-3x+2=0; x⁴-7x²+6x=0 . Z góry dzieki

Answer 1

x³-3√2x²+√2x-6=0

x²(x-3√2)+√2(x-3√2)=0

(x-3√2)(x²+√2)=0

x-3√2=0 lubx²+√2=0

x=3√2

 

x³-5x-4=0

x³-x-4x-4=0

x(x²-1)-4(x-1)=0

(x(x-1)(x+1)-4(x-1)=0

(x-1)(x(x+1)-4)=0

x-1=0 lub x(x+1)-4=0

x=1 lub x²+x-4=0

delta = 17

√delta = √17

x1=(-1-√17)/2

x2=(-1+√17)/2

zatem x=1 lub x=(-1-√17)/2 lub x=(-1+√17)/2

 

x³-3x+2=0

x³-x-2x+2=0

x(x²-1)-2(x-1)=0

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

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

x-1=0 lub x(x+1)-2 = 0

x=1 lub x²+x-2=0

x=1 lub delta = 9

√delta=3

x1=-1-3/2=-4/2=-2

x2=-1+3/2=2/2=1

zatem x=1 lub x=-2

 

x⁴-7x²+6x=0

x⁴-x²-6x²+6x=0

x²(x²-1)-6x(x-1)=0

x²(x-1)(x+1)-6x(x-1)=0

(x-1)(x²(x+1)-6x)=0

x-1=0 lub (x²(x+1)-6x)=0

x=1 lub x³-x²-6x=0

x(x²- x-6)=0

x=0 lub  (x²- x-6=0

delta = 25

√delta = 5

x1=1-5/2=-6/2=-3

x2=-1+5/2=4/2=2

zatem x=0 lub x=1 lub x=2 lub x=-3

Answer 2

x³-3√2x²+√2x-6=0

x²(x-3√2)+√2(x-3√2)=0

(x-3√2)(x²+√2)=0

 

x-3√2=0

x=3√2

 

x²+√2=0

x²=-√2 <-- brak rozwiązań

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1-x³=x²-x

-x³-x²+x+1=0

-x²(x+1)+1(x+1)=0

(-x²+1)(x+1)=0

 

-x²+1=0

-x²=-1

x²=1

 

x=-1 ∨ x=1

 

x+1=0

x=-1

-------------------------

 

x³-5x-4=0

x³-x-4x-4=0

x(x²-1)-4(x+1)=0

x(x+1)(x-1)-4(x+1)=0

(x+1)(x(x-1)-4)=0

(x+1)(x²-x-4)=0

 

x+1=0

x=-1

 

Δ=(-1)²-4*1*(-4)

Δ=1+16

Δ=17

√Δ=√17

 

x₁=(-(-1)-√17)/(2*1)

x₁=(1-√17)/2

 

x₂=(-(-1)+√17)/(2*1)

x₂=(1+√17)/2

 

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x³-3x+2=0

x³-x-2x+2=0

x(x²-1)-2(x-1)=0

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

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

(x-1)(x²+x-2)=0

 

x-1=0

x=1

 

Δ=1²-4*1*(-2)

Δ=1+8

Δ=9

√Δ=3

 

x₁=(-1-3)/(2*1)

x₁=-4/2

x₁=-2

 

x₂=(-1+3)/(2*1)

x₂=2/2

x₂=1

-------------------------

 

x⁴-7x²+6x=0

x(x³-7x+6)=0

x(x³-x-6x+6)=0

x(x(x²-1)-6(x-1))=0

x(x(x-1)(x+1)-6(x-1))=0

x(x-1)(x(x+1)-6)=0

x(x-1)(x²+x-6)=0

 

x=0

 

x-1=0

x=1

 

Δ=1²-4*1*(-6)

Δ=1+24

Δ=25

√Δ=5

 

x₁=(-1-5)/(2*1)

x₁=6/2

x₁=-3

 

x₂=(-1+5)/(2*1)

x₂=4/2

x₂=2