Zadanie 3.80 na str. 82 w zbiorze zadań "Matematyka" Kurczab, Świda.Skróć ułamki. Podaj konieczne za.... Question from @Dominika1547

Dominika1547 @Dominika1547

October 2018 1 202 Report

Zadanie 3.80 na str. 82 w zbiorze zadań "Matematyka" Kurczab, Świda.

Skróć ułamki. Podaj konieczne założenia. Daję Naj!!!

a) x^2-4

x^2+4x+4

b) 9+6x+x^2

x^2-9

c) x^2-5x

(x-5)^2

d) 6x-2

9x^2-6x+1

e) (x-1)^2+4x

x^2-1

f) (x+4)^2-16x

(x-4)^2

g) 4-(x-2)^2

16x-8x^2+x^3

h) x^3+10x^2+25x

[1-(x+4)^2]X

Z góry dziękuję ;)

Roma A)
$\frac{x^2-4}{x^2+4x+4}=\\\\ Dziedzina:\\\ x^2+4x+4\neq0\\\ 
\Delta=4^2-4\cdot1\cdot4=0\\\ x_0=\frac{-4}{2\cdot1}=-2 \\\ D = \{x \in 
R: x\neq-2 \} \\\\ 
=\frac{(x-2)(x+2)}{(x+2)^2}=\frac{(x-2)(x+2)}{(x+2)(x+2)}=\frac{x-2}{x+2}$

b)
$\frac{9+6x+x^2}{x^2-9}=\\\\
 Dziedzina:\\\ x^2-9\neq0\\\ (x-3)(x+3)=0 \\\ x-3=0\ \vee \ x+3=0\\\ x=3
 \ \vee \ x=-3 \\\ D=\{x \in R: x\neq -3 \ i \ x\neq 3 \} \\\\ 
=\frac{(x+3)^2}{(x-3)(x+3)}=\frac{(x+3)(x+3)}{(x-3)(x+3)}=\frac{x+3}{x-3}$

c)
$\frac{x^2-5x}{(x-5)^2}
 =\\\\ Dziedzina:\\\ (x-5)^2\neq0\\\ x-5=0 \\\ x=5 \\\ D = \{x \in R: 
x\neq5 \} \\\\ 
=\frac{x(x-5)}{(x-5)^2}=\frac{x(x-5)}{(x-5)(x-5)}=\frac{x}{x-5}$

d)
$\frac{6x-2}{9x^2-6x+1}=\\\\
 Dziedzina:\\\ 9x^2-6x+1\neq0 \\\ (3x-1)^2=0\\\ 3x-1=0\\\ 3x=1 \ /:3 \\\
 x= \frac{1}{3} \\\ D = \{x \in R: x\neq \frac{1}{3} \} \\\\ 
=\frac{2(3x-1)}{(3x-1)^2}=\frac{2(3x-1)}{(3x-1)(3x-1)}=\frac{2}{3x-1}$

e)
$\frac{(x-1)^2+4x}{x^2-1}=\\\\
 Dziedzina:\\\ x^2-1\neq0\\ (x-1)(x+1)=0 \\\ x- 1 = 0 \ \vee \ x + 1 = 0
 \\\ x =1 \ \vee \ x =-1 \\\ D = \{x \in R: x\neq-1 \ i \ x \neq 1 
\}\\\\ =\frac{(x-1)^2+4x}{(x-1)(x+1)}=\frac{x^2 -2x 
+1+4x}{(x-1)(x+1)}=\frac{x^2+2x +1}{(x-1)(x+1)}=\frac{(x 
+1)^2}{(x-1)(x+1)}= \frac{(x+1)(x+1)}{(x-1)(x+1)} =\\\ =\frac{x +1}{x-1}$

f)
$\frac{(x+4)^2-16x}{(x-4)^2}=\\\\
 Dziedzina:\\\ (x-4)^2\neq0\\\ x-4 =0 \\\ x =4 \\\ D = \{x \in R: x \neq
 4 \}\\\\ 
=\frac{x^2+8x+16-16x}{(x-4)^2}=\frac{x^2-8x+16}{(x-4)^2}=\frac{(x-4)^2}{(x-4)^2}=1$

g)
$\frac{4-(x-2)^2}{16x-8x^2+x^3}=\\\\
 Dziedzina:\\\ 16x-8x^2+x^3\neq0\\\ 16x-8x^2+x^3=0 \\\ 
x(x^2-8x+16)=0\\\ x(x-4)^2=0\\\ x = 0 \ \vee \ x - 4 = 0 \\\ x=0\ \vee \
 x=4 \\\ D = \{x \in R: x\neq 0 \ i \ x \neq 4 \}\\\\ 
=\frac{4-x^2+4x-4}{x(x-4)^2}=\frac{-x^2+4x}{x(x-4)^2}=\frac{-x(x-4)}{x(x-4)(x-4)}=\frac{-1}{x-4}=-\frac{1}{x-4}=\frac{1}{4-x}$

h)
$\frac{x^3+10x^2+25x}{(1-(x+4)^2)x}=\\\\ Dziedzina:\\\ (1-(x+4)^2))x\neq0 \\\ (1-(x+4)^2))x=0 \\\ 
(1-x^2-8x-16)x=0 \\\ (-x^2-8x-15)x = 0 \\\ -x^2-8x-15 = 0 \ \vee \ x = 0
 \\\ \Delta=(-8)^2-4\cdot(-1)\cdot(-15)=64-60=4; \ \sqrt{\Delta} = 
\sqrt{4}=2 \\\ x_1=\frac{8-2}{2\cdot(-1)}= \frac{6}{-2} =-3 \\\ 
x_2=\frac{8+2}{2\cdot(-1)}= \frac{10}{-2} =-5 \\\ x = -3 \ \vee \ x = - 5 \ \vee \ x = 0 \\\ D = \{x \in R: x\neq -5 \ i \ x \neq -3 \ i \ x \neq 0 \}$

$=\frac{x(x^2+10x+25)}{(1-(x+4)^2)x}=\frac{x(x+5)^2}{(1-(x+4)^2)x}=\frac{(x+5)^2}{1-(x+4)^2}
 =\frac{(x+5)(x+5)}{1^2-(x+4)^2} =\\\ =\frac{(x+5)(x+5)}{(1-(x+4))(1+(x+4))}
 =\frac{(x+5)(x+5)}{(1-x-4)(1+x+4)} =\frac{(x+5)(x+5)}{(-x-3)(x+5)} =\\\
 =\frac{x+5}{-x-3}=-\frac{x+5}{x+3}$

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