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Bardzo proszę o szybką pomoc! do zdobycia 90pkt <3



Bardzo Proszę O Szybką Pomoc Do Zdobycia 90pkt Lt3 class=

Odpowiedź :

Warcholm4

Odpowiedź:

a)

[tex]\frac{3}{x-2} + \frac{x+1}{x+2} - \frac{x^{2}+4 }{(x-2)*(x+2)}=\frac{3(x+2)+(x-2) *(x+1)-(x^{2} +4) }{(x-2)*(x+2)} = \frac{3x+6+x^{2}+x -2x-2-x^{2} -4}{x^{2}-4 } = \frac{2x}{x^{2} -4}[/tex]

x = 2, x = -2

b)

[tex]\frac{4}{x*(x+6)} -\frac{1-x}{2x}+ \frac{x-1}{x+6} = \frac{8-(x+6)*(1-x)+2x*(x-1)}{2x*(x+6)} =\frac{8-(x-x^{2} +6-6x)+2x^{2} -2x}{2x^{2} +12x} = \frac{8-(-5x-x^{2} +6)+2x^{2} -2x}{2x^{2} +12x} =\frac{8+5x+x^{2} -6+2x^{2} -2x}{2x^{2} +12x} = \frac{3x^{2} +3x+2}{2x^{2} +12x}[/tex]

x = 0, x = -6

c)[tex]\frac{x^{2} }{(2x-3)*(2x+3)} +\frac{2-x}{2x-3} -\frac{6}{-(2x-3)}= \frac{x^{2} }{(2x-3)*(2x+3)} +\frac{2-x}{2x-3} +\frac{6}{2x-3}= \frac{x^{2} +(2x+3)*(2-x)+6(2x+3)}{(2x-3)*(2x+3)} =\frac{x^{2} +4x-2x^{2} +6-3x+12x+18}{4x^{2} -9} =\frac{-x^{2} +13x+24}{4x^{2} -9}[/tex]x = -1,5, x = 1,5

d)

[tex]\frac{2x^{2} -7}{(3x-1)^{2} } - \frac{2-3x^{2} }{3x-1} -x=\frac{2x^{2} -7-(3x-1)*(2-3x^{2} )-(3x-1)^{2} *x}{(3x-1)^{2} } = \frac{2x^{2} -7-(6x-9x^{3} -2+3x^{2} )-(9x^{2} -6x+1)x}{(3x-1)^{2}} = \frac{2x^{2} -7-6x+9x^{3} +2-3x^{2} -(9x^{3} -6x^{2} +x)}{(3x-1)^{2}} =\frac{-x^{2} -5-6x+9x^{3} -9x^{3} +6x^{2} -x}{(3x-1)^{2}} =\frac{5x^{2} -7x-5}{(3x-1)^{2}}[/tex]

x = 1/3

Krysia

[tex]a)x-2\neq 0\ \ \vee  \ \ x+2\neq 0\\\\x\neq 2\ \ \vee \ \ x\neq -2\\\\D=R\setminus \left \{ -2,2 \right \}\\\\\frac{3}{ x-2} +\frac{ x+1}{x+2} -\frac{ x^2+4}{x^2-4}= \frac{3}{ x-2} +\frac{ x+1}{x+2} -\frac{ x^2+4 }{(x-2)(x+2) } =\\\\\\=\frac{3(x+2)+(x+1)(x-2)-(x^2+4)}{(x-2)(x+2)} = \frac{3 x+6+x^2-2x+x-2 - x^2-4 }{(x-2)(x+2)} =\\\\= \frac{2x }{(x-2)(x+2)}[/tex]

[tex]b)\\\\x^2+6x\neq 0\\\\x(x+6)\neq 0\\\\x\neq 0\ \ \vee \ \ x+6\neq 0\\\\x\neq 0\ \ \vee \ \ x \neq -6\\\\D=R\setminus \left \{ -6,0 \right \}[/tex]

[tex]\frac{4}{x^2+6x}-\frac{1-x}{2x}+\frac{x-1}{x+6}=\frac{4}{x(x +6) }-\frac{1-x}{2x}+\frac{x-1}{x+6}=\frac{4*2-(1-x)*(x+6) +(x-1)*2x }{2x(x+6)} =\\\\=\frac{8-(x+6-x^2-6x) +2x^2-2x}{2x(x+6)} =\frac{8- x-6+x^2+6x +2x^2-2x}{2x(x+6)}=\\\\=\frac{3x^2+3x+2}{2x(x+6)}[/tex]

[tex]c)\\\\4x^2-9\neq 0\\\\(2x-3)(2x+3)\neq 0 \\\\2x-3\neq 0\ \ \vee \ \ 2x+3\neq 0\\\\2x \neq 3\ \ \vee \ \ 2x \neq -3 \\\\ x \neq \frac{3}{2}\ \ \vee \ \ x \neq -\frac{3}{2}\\\\D=R\setminus \left \{ -\frac{3}{2},\frac{3}{2} \right \}[/tex]

[tex]\frac{x^2}{4x^2-9} + \frac{2-x}{2x-3}-\frac{6}{3-2x}= \frac{x^2}{(2x-3)(2x+3)} + \frac{2-x}{2x-3}+\frac{6}{ 2x-3} =\\\\\\=\frac{x^2+(2-x)(2x+3)+6(2x+3)}{(2x-3)(2x+3)}=\frac{x^2+4x+6-2x^2-3x +12x+18}{(2x-3)(2x+3)}=\\\\ =\frac{-x^2+13x+ 24}{(2x-3)(2x+3)}[/tex]

[tex]d)\\\\3x-1\neq 0\\\\3x\neq 1\ \ |:3\\\\x\neq \frac{1}{3}\\\\D=R\setminus \left \{ \frac{1}{3} \right \}[/tex]

[tex]\frac{2x^2-7}{ 9x^2-6x+1 } - \frac{2-3x^2 }{3x-1} - x =\frac{2x^2-7}{ (3x-1)^2 } - \frac{2-3x^2 }{3x-1} - x = \frac{ 2x^2-7 -(2-3x^2)(3x-1)-x(3x-1)^2}{(3x-1)^2}=\\\\ = \frac{ 2x^2-7 -(6x-2-9x^3+3x^2) -x(9x^2-6x+1) }{(3x-1)^2} = \frac{ 2x^2-7 - 6x+2+9x^3-3x^2 -9x^3+6x^2-x }{(3x-1)^2} =\\\\ = \frac{ 5x^2-7x-5}{(3x-1)^2}[/tex]