Wykonaj działania ...- Zadanie 22: Matematyka poznać, zrozumieć 2. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

a) Zał:

$x - 1 \neq = 0, x + 1 \neq = 0, x \neq = 0$

$x \neq = 1, x \neq = - 1, x \neq = 0$

$\frac{1}{x - 1} : (\frac{1}{( x - 1 ) ^{2}} + \frac{1}{x ^{2} - 1}) = \frac{1}{x - 1} : (\frac{1}{( x - 1 ) ^{2}} + \frac{1}{( x - 1 ) ( x + 1 )}) =$

$= \frac{1}{x - 1} : (\frac{x + 1}{( x - 1 ) ^{2} ( x + 1 )} + \frac{x - 1}{( x - 1 ) ^{2} ( x + 1 )}) = \frac{1}{x - 1} : \frac{x + 1 + x - 1}{( x - 1 ) ^{2} ( x + 1 )} =$

$= \frac{1}{x - 1} : \frac{2 x}{( x - 1 ) ^{2} ( x + 1 )} = \frac{1}{x - 1} \cdot \frac{( x - 1 ) ^{2} ( x + 1 )}{2 x} = \frac{( x - 1 ) ( x + 1 )}{2 x} = \frac{x ^{2} - 1}{2 x}$

b) Zał:

$x - 4 \neq = 0, x + 4 \neq = 0, x - 12 \neq = 0$

$x \neq = 4, x \neq = - 4, x \neq = 12$

$(\frac{2}{x - 4} - \frac{4}{x + 4}) : (\frac{x ^{2} - 16}{2 x + 8} - \frac{x}{4} - 1) =$

$= (\frac{2 ( x + 4 )}{( x - 4 ) ( x + 4 )} - \frac{4 ( x - 4 )}{( x - 4 ) ( x + 4 )}) : (\frac{( x - 4 ) ( x + 4 )}{2 ( x + 4 )} - \frac{x}{4} - 1) =$

$= (\frac{2 x + 8}{( x - 4 ) ( x + 4 )} - \frac{4 x - 16}{( x - 4 ) ( x + 4 )}) : (\frac{2 ( x - 4 )}{4} - \frac{x}{4} - \frac{4}{4}) =$

$= \frac{2 x + 8 - 4 x + 16}{( x - 4 ) ( x + 4 )} : \frac{2 x - 8 - x - 4}{4} = \frac{- 2 x + 24}{( x - 4 ) ( x + 4 )} \cdot \frac{4}{x - 12} =$

$= \frac{- 2 ( x + 12 )}{( x - 4 ) ( x + 4 )} \cdot \frac{4}{x - 12} = \frac{- 8}{x ^{2} - 16} = \frac{8}{16 - x ^{2}}$

c) Zał:

$x - 5 \neq = 0, 2 x - 1 \neq = 0, 2 x - 3 \neq = 0, 2 x + 3 \neq = 0, x \neq = 0$

$x \neq = 5, x \neq = \frac{1}{2}, x \neq = \frac{3}{2}, x \neq = - \frac{3}{2}, x \neq = 0$

$(\frac{\frac{2 x ^{2} - 7 x - 15}{4 x ^{2} - 9}}{125 - x ^{3}} - \frac{3}{\frac{x ^{2} + 5 x + 25}{2 x - 1}}) \cdot (\frac{1}{2} + \frac{5}{2 x} + \frac{25}{2 x ^{2}}) =$

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

$= (\frac{- 1}{( x ^{2} + 5 x + 25 ) ( 2 x - 3 )} - \frac{( 6 x - 3 ) ( 2 x - 3 )}{( x ^{2} + 5 x + 25 ) ( 2 x - 3 )} \cdot \frac{x ^{2} + 5 x + 25}{2 x ^{2}} =$

$= (\frac{- 1}{( x ^{2} + 5 x + 25 ) ( 2 x - 3 )} - \frac{( 12 x ^{2} - 18 x - 6 x + 9 )}{( x ^{2} + 5 x + 25 ) ( 2 x - 3 )}) \cdot \frac{x ^{2} + 5 x + 25}{2 x ^{2}} =$

$= \frac{- 1 - 12 x ^{2} + 18 x + 6 x - 9}{( x ^{2} + 5 x + 25 ) ( 2 x - 3 )} \cdot \frac{x ^{2} + 5 x + 25}{2 x ^{2}} = \frac{- 12 x ^{2} + 24 x - 10}{2 x - 3} \cdot \frac{1}{2 x ^{2}} =$

$= \frac{- 6 x ^{2} + 12 x - 5}{2 x ^{2} ( 2 x - 3 )}$

d) Zał:

$x - 4 \neq = 0, x + 4 \neq = 0, x - 2 \neq = 0, x + 2 \neq = 0$

$x \neq = 4, x \neq = - 4, x \neq = 2, x \neq = - 2$

$(\frac{1}{16 - x ^{2}} - \frac{1}{x ^{2} - 4 x + 16}) : \frac{x ^{2} - 4}{x - 4} =$

$= (\frac{1}{( 4 - x ) ( 4 + x )} - \frac{1}{x ^{2} - 4 x + 16}) \cdot \frac{x - 4}{( x - 2 ) ( x + 2 )} =$

$= (\frac{x ^{2} - 4 x + 16}{( 4 - x ) ( 4 + x ) ( x ^{2} - 4 x + 16 )} - \frac{( 4 - x ) ( 4 + x )}{( 4 - x ) ( 4 + x ) ( x ^{2} - 4 x + 16 )}) \cdot \frac{x - 4}{( x - 2 ) ( x + 2 )} =$

$= \frac{x ^{2} - 4 x + 16 - 16 + x ^{2}}{( 4 - x ) ( 4 + x ) ( x ^{2} - 4 x + 16 )} \cdot \frac{x - 4}{( x - 2 ) ( x + 2 )} =$

$= \frac{2 x ^{2} - 4 x}{- ( x - 4 ) ( x + 4 ) ( x ^{2} - 4 x + 16 )} \cdot \frac{x - 4}{( x - 2 ) ( x + 2 )} =$

$= \frac{2 x ( x - 2 )}{- ( x + 4 ) ( x ^{2} - 4 x + 16 )} \cdot \frac{1}{( x - 2 ) ( x + 2 )} = \frac{2 x}{- ( x + 4 ) ( x ^{2} - 4 x + 16 )} \cdot \frac{1}{x + 2} =$

$= \frac{- 2 x}{( x + 4 ) ( x ^{2} - 4 x + 16 ) ( x + 2 )} = \frac{- 2 x}{( x ^{3} + 64 ) ( x + 2 )}$

e) Zał:

$x \neq = 0, x + 2 \neq = 0, x - 2 \neq = 0, x + 4 \neq = 0, x - 4 \neq = 0$

$x \neq = 0, x \neq = - 2, x \neq = 2, x \neq = - 4, x \neq = 4$

$(\frac{4}{x ^{2} + 2 x} - \frac{x}{x + 2}) \cdot \frac{4 x - 8}{x ^{2} + 2 x - 8} : \frac{x ^{2} - 6 x + 8}{x ^{2} - 4} =$

$= (\frac{4}{x ( x + 2 )} - \frac{x}{x + 2}) \cdot \frac{4 ( x - 2 )}{( x + 4 ) ( x - 2 )} : \frac{( x - 2 ) ( x - 4 )}{( x - 2 ) ( x + 2 )} =$

$= (\frac{4}{x ( x + 2 )} - \frac{x ^{2}}{x ( x + 2 )}) \cdot \frac{4}{x + 4} \cdot \frac{x + 2}{x - 4} = \frac{4 - x ^{2}}{x ( x + 2 )} \cdot \frac{4}{x + 4} \cdot \frac{x + 2}{x - 4} = \frac{4 ( 4 - x ^{2} )}{x ( x ^{2} - 16 )}$

f) Zał:

$2 x - 3 \neq = 0, 2 x + 3 \neq = 0, 16 x^{3} \neq = 0$

$x \neq = \frac{3}{2}, x \neq = - \frac{3}{2}, x \neq = 0$

$(4 x^{2} + 9) (\frac{1}{4 x ^{2} - 9} - \frac{1}{16 x ^{4} - 81}) : (\frac{1}{8 x ^{3} - 27} + \frac{1}{8 x ^{3} + 27}) (\frac{1}{4 x ^{2} - 6 x + 9} - \frac{1}{4 x ^{2} + 6 x + 9}) =$

$= (4 x^{2} + 9) (\frac{1}{4 x ^{2} - 9} - \frac{1}{( 4 x ^{2} - 9 ) ( 4 x ^{2} + 9 )}) : (\frac{8 x ^{3} + 27}{( 8 x ^{3} - 27 ) ( 8 x ^{3} + 27 )} + \frac{8 x ^{3} - 27}{( 8 x ^{3} - 27 ) ( 8 x ^{3} + 27 )}) (\frac{4 x ^{2} + 6 x + 9}{( 4 x ^{2} - 6 x + 9 ) ( 4 x ^{2} + 6 x + 9 )} - \frac{4 x ^{2} - 6 x + 9}{( 4 x ^{2} - 6 x + 9 ) ( 4 x ^{2} + 6 x + 9 )}) =$

$= (4 x^{2} + 9) (\frac{4 x ^{2} + 9}{( 4 x ^{2} - 9 ) ( 4 x ^{2} + 9 )} - \frac{1}{( 4 x ^{2} - 9 ) ( 4 x ^{2} + 9 )}) : \frac{8 x ^{3} + 27 + 8 x ^{3} - 27}{( 8 x ^{3} - 27 ) ( 8 x ^{3} + 27 )} \cdot \frac{4 x ^{2} + 6 x + 9 - 4 x ^{2} + 6 x - 9}{( 4 x ^{2} - 6 x + 9 ) ( 4 x ^{2} + 6 x + 9 )} =$

$= (4 x^{2} + 9) \cdot \frac{4 x ^{2} + 9 - 1}{( 4 x ^{2} - 9 ) ( 4 x ^{2} + 9 )} \cdot \frac{( 8 x ^{3} - 27 ) ( 8 x ^{3} + 27 )}{16 x ^{3}} \cdot \frac{12 x}{( 4 x ^{2} - 6 x + 9 ) ( 4 x ^{2} + 6 x + 9 )} =$

$= \frac{4 x ^{2} + 8}{4 x ^{2} - 9} \cdot \frac{( 2 x - 3 ) ( 4 x ^{2} + 6 x + 9 ) ( 2 x + 3 ) ( 4 x ^{2} - 6 x + 9 )}{16 x ^{3}} \cdot \frac{12 x}{( 4 x ^{2} - 6 x + 9 ) ( 4 x ^{2} + 6 x + 9 )} =$

$= \frac{4 ( x ^{2} + 2 )}{( 2 x - 3 ) ( 2 x + 3 )} \cdot \frac{( 2 x - 3 ) ( 2 x + 3 )}{16 x ^{3}} \cdot \frac{12 x}{1} = \frac{4 ( x ^{2} + 2 ) \cdot 12 x}{16 x ^{3}} = \frac{3 ( x ^{2} + 2 )}{x ^{2}}$

Magdalena Matusik

Nauczycielka matematyki

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