Question

Rozwiąż równania: 1 ) -x^4+7x^2=3(x^4+1) 2) (5x-8)(4_3x)+x(4-x)=0 3 ) (2x-3)(x+5)=(5x+9)(6+3x) 4) x(2x+7)-5=(x+4)(5-x) 5) (3x-1)(2-x)-x(x+3)=2 Proszę o pomoc ponieważ nie mam pojęcia jak się do tego zabrać

Answer 1

1) -x^4+7x^2=3(x^4+1)

-x⁴+7x²=3x⁴+3

4x⁴-7x²+3=0

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Podstawienie:

x²=t

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4t²-7t+3=0

Δ=b²-4ac

Δ=(-7)²-4*4*3=49-48=1

√Δ=1

t₁=[-b-√Δ]/2a=[7-1]/8=6/8=3/4

t₂=[-b+√Δ]/2a=[7+1]/8=8/8=1

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Wracam do podstawienia:

x²=t₁                 lub                 x²=t₂

x²=3/4                                    x²=1

x²-3/4=0                                x²-1=0

(x - √3/2)(x + √3/2)=0          (x-1)(x+1)=0

x₁=√3/2  lub  x₂=-√3/2  lub  x₃=1  lub  x₄=-1

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2) (5x-8)(4-3x)+x(4-x)=0

20x-15x²-32+24x+4x-x²=0

-16x²+48x-32=0         |:(-16)

x²-3x+2=0

Δ=(-3)²-4*1*2=9-8=1

√Δ=1

x₁=[3-1]/2=2/2=1

x₂=[3+1]/2=4/2=2

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3) (2x-3)(x+5)=(5x+9)(6+3x)

2x²+10x-3x-15=30x+15x²+54+27x

13x²+50x+69=0

Δ=50²-4*13*69=2500-3588=-1088

Δ<0  - brak pierwiastków

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4) x(2x+7)-5=(x+4)(5-x)

2x²+7x-5=5x-x²+20-4x

3x²+6x-25=0

Δ=6²-4*3*(-25)=36+300=336

√Δ=√336=4√21

x₁=[-6-4√21]/6=-(3+2√21)/3

x₂=[-6+4√21]/6=-(3-2√21)/3

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5) (3x-1)(2-x)-x(x+3)=2

6x-3x²-2+x-x²-3x-2=0

-4x²+4x-4=0

-4(x²-1+1)=0

Δ=(-1)²-4*1*1=1-4=-3

Δ<0  - brak pierwiastków