Rozwiąż równanie......

[tex]a)\frac{x^2+3x}{2} -\frac{x^2+7}{3} =\frac{3x+2}{3} |*6\\3(x^2+3x)-2(x^2+7)=2(3x+2)\\3x^2+9x-2x^2-14-2(3x+2)=0\\x^2+9x-14-6x-4=0\\x^2+3x-18=0[/tex]

∆=3²-4*1*(-18)=9+72=81

√∆=9

[tex]x_1=\frac{-3-9}{2} =-6\\x_2=\frac{-3+9}{2} =3[/tex]

x∈{-6;3}

[tex]b)\frac{x^2-3}{8} -\frac{x^2}{4}=\frac{x^2+6}{2} +x^2-5|*8\\x^2-3-2x^2=4(x^2+6)+8x^2-40\\-x^2-3-4(x^2+6)-8x^2+40=0\\-9x^2+37-4x^2-24=0\\-13x^2+13=0\\-13x^2=-13\\x^2=1\\|x|=1\\x=1 lub x=-1[/tex]

x∈{-1;1}

[tex]c)x(x+7)-5(x-7)(x-6)=6(2x+3)^2-320\\x^2+7x-5(x^2-6x-7x+42)-6(4x^2+12x+9)+320=0\\x^2+7x-5(x^2-13x+42)-6(4x^2+12x+9)+320=0\\x^2+7x-5x^3+65x-210-24x^2-72x-54+320=0\\-28x^2+56=0\\-28x^2=-56\\x^2=2[/tex]

[tex]|x|=\sqrt{2} \\x=\sqrt{2} lub x=-\sqrt{2}[/tex]

x∈{-√2;√2}

[tex]d)x((x-1)(x+1)+x(1-x))=x((x-1)(x+1)-x^2)\\x((x-1)(x+1)+x(1-x))-x((x-1)(x+1)-x^2)=0\\x(x^2-1+x-x^2)-x(x^2-1-x^2)=0\\x(-1+x)+1=0\\x^2-x+1=0[/tex]

∆=1-4*1*1=1-4=-3

Brak rozwiązania, bo delta mniejsza od zera

123bodzio 123bodzio · Answer 2 · 2021-09-09T13:20:34+02:00

Odpowiedź:

zad 11

a)

(x² + 3x)/2 - (x² + 7)/3 = (3x + 2)/3 | * 6

3(x² + 3x) - 2(x² + 7) = 2(3x + 2)

3x² + 9x - 2x² - 14 = 6x + 4

x² + 9x - 14 = 6x + 4

x² + 9x - 6x - 14 - 4 = 0

x² + 3x - 18 = 0

a = 1 , b = 3 , c = - 18

Δ = b² - 4ac = 3² - 4 * 1 * (- 18) = 9 + 72 = 81

√Δ = √81 = 9

x₁ = ( - b - √Δ)/2a = (- 3 - 9)/2 = - 12/2 = - 6

x₂ = (- b + √Δ)/2a = (- 3 + 9)/2 = 6/2 = 3

b)

(x² - 3)/8 - x²/4 = (x² + 6)/2 + x² - 5 | * 8

x² - 3 - 2x² = 4(x² + 6) + 8x² - 40

- x² - 3 = 4x² + 24 + 8x² - 40

- x² - 3 = 12x² - 16

- x² - 12x² = - 16 + 3

- 13x² = - 13 | : (- 13)

x² = 1

x² - 1 = 0

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

x - 1 = 0 ∨ x + 1 = 0

x = 1 ∨ x = - 1

c)

x(x + 7) - 5(x - 7)(x - 6) = 6(2x + 3)² - 320

x² + 7x - 5(x² - 7x - 6x + 42) = 6(4x² + 12x + 9) - 320

x² + 7x - 5(x² - 13x + 42) = 24x² + 72x + 54 - 320

x² + 7x - 5x² + 65x - 210 = 24x² + 72x - 266

- 4x² + 72x - 210 = 24x² + 72x - 266

- 4x² - 24x² + 72x - 72x = - 266 + 210

- 28x² = - 56 | : (- 28)

x² = 2

x² - 2 = 0

(x - √2)(x + √2) = 0

x - √2 = 0 ∨ x + √2 = 0

x = √2 ∨ x = - √2

d)

x[(x - 1)(x + 1) + x(1 - x)] = x[(x - 1)(x + 1) - x²]

x(x² + x - x²) = x(x² - 1 - x²)

x * x = x * (- 1)

x² = - x

x² + x = 0

x(x + 1) = 0

x = 0 ∨ x + 1 = 0

x = 0 ∨ x = - 1