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3 years ago

Pls do all of them correctly <3

Mathematics

2 answers:

inessss [21]3 years ago

7 0

#12; 2•3•5^2

#13; -95

#14; -83

AfilCa [17]3 years ago

7 0

Answer:

#12; 2•3•5^2

#13; -95

#14; -83p

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creo que es esto :)

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hope this helped

stay safe

brainliest is appreciated only 1more to level up please help :)))

8 0

4 years ago

(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions

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$\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}$
$\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}$
$\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}$
$\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}$
$\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}$
$\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0$
$\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0$

this will be continued

5 0

4 years ago