Question

Zadanie 1. Rozwiąż równania
[tex]a) \ 2x^{2}-10x+12 = 0\\\\b) \ x^{2}-6x+9 = 0\\\\c) \ \frac{1}{x^{3}+x^{2}+1} = 1\\\\d) \ \frac{1}{x+2}-\frac{6}{x^{2}-4} = 0[/tex]

Answer 1

Rozwiązanie w załączniku

Answer 2

Verified answer

a)

[tex]2x^2-10x+12=0\\a=2, b=-10, c=12\\\Delta=(-10)^2-4*2*12=100-96=4\\\sqrt{\Delta}=2\\[/tex]

Równanie ma dwa rozwiązania:

[tex]x_1=\dfrac{10-2}4=\dfrac84=2\\\\x_2=\dfrac{10+2}4=\dfrac{12}4=3\\\\\boxed{x=2 \vee x=3}[/tex]

b)

[tex]x^2-6x+9=0\\a=1, b=-6, c=9\\\\\Delta=(-6)^2-4*1*9=36-36=0[/tex]

Równanie ma jedno rozwiązanie:

[tex]x_0=\dfrac{6}2=3\\\\\boxed{x=3}[/tex]

c)

[tex]\dfrac{1}{x^3+x^2+1}=1\\\\x^3+x^2+1 \neq 0[/tex]

To równanie bedzie prawdziwe tylko, jeżeli mianownik będzie równy 1

[tex]x^3+x^2+1=0 /-1\\x^3+x^2=0\\x^2(x+1)=0\\\begin{array}{c c c}x^2=0 & \vee & x+1 =0 \\x=0 & \vee & x=-1\end{array}[/tex]

[tex]\boxed{x=0 \vee x=-1}[/tex]

d)

[tex]\dfrac1{x+2}-\dfrac6{x^2-4}=0\\\\x+2 \neq 0 \wedge x^2-4 \neq 0\\x \neq -2 \wedge x^2 \neq 4\\x \neq -2 \wedge x \neq 2\\D: x\in \mathbb{R} \setminus \{-2, 2\}[/tex]

[tex]\dfrac{x^2-4}{(x+2)(x^2-4)}-\dfrac{6(x+2)}{(x+2)(x^2-4)}=0\\\\\dfrac{x^2-4-6(x+2)}{(x+2)(x+2)(x-2)}=0 \\\\\dfrac{x^2-4-6x-12}{(x+2)(x+2)(x-2)}=0\\\\\dfrac{x^2-6x-16}{(x+2)(x+2)(x-2)}= 0 /*[(x+2)(x+2)(x-2)]\\\\x^2-6x-16=0\\a=1, b=-6, c=-16\\\Delta=(-6)^2-4*1*(-16)=36+64=100\\\sqrt{\Delta}=10\\\\x_1=\dfrac{6-10}2=\dfrac{-4}2=-2 \notin D\\\\x_2=\dfrac{6+10}2=\dfrac{16}2=8\\\\\boxed{x=8}[/tex]