Wykonaj działania- Zadanie 3: MATeMAtyka 2. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

$a)$

$za ł o \overset{z}{˙} enia:$

${x^{2} + 4 x \neq = 0 x^{2} - 4 x \neq = 0$

${x (x + 4) \neq = 0 x (x - 4) \neq = 0$

${x \neq = 0 i x \neq = - 4 x \neq = 0 i x \neq = 4$

$D = R \ {- 4, 0, 4}$

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

$= \frac{3 x - 12 - x ^{2} + 4 x}{x ( x + 4 ) ( x - 4 )} - \frac{3 x + 12 + x ^{2} + 4 x}{x ( x + 4 ) ( x - 4 )} = \frac{- x ^{2} + 7 x - 12}{x ( x + 4 ) ( x - 4 )} - \frac{x ^{2} + 7 x + 12}{x ( x + 4 ) ( x - 4 )} =$

$= \frac{- x ^{2} + 7 x - 12 - x ^{2} - 7 x - 12}{x ( x + 4 ) ( x - 4 )} = \frac{- 2 x ^{2} - 24}{x ( x + 4 ) ( x - 4 )}$

$b)$

$za ł o \overset{z}{˙} enia:$

$⎩ ⎨ ⎧ 4 x^{2} - 1 \neq = 0 ∣ : 4 2 x^{2} - x \neq = 0 ∣ : 2 2 x + 1 \neq = 0 ∣ - 1$

$⎩ ⎨ ⎧ x^{2} - \frac{1}{4} \neq = 0 x^{2} - \frac{1}{2} x \neq = 0 2 x \neq = - 1 ∣ : 2$

$⎩ ⎨ ⎧ (x - \frac{1}{2}) (x + \frac{1}{2}) \neq = 0 x (x - \frac{1}{2}) \neq = 0 x \neq = - \frac{1}{2}$

$⎩ ⎨ ⎧ x \neq = \frac{1}{2} i x \neq = - \frac{1}{2} x \neq = 0 i x \neq = \frac{1}{2} x \neq = - \frac{1}{2}$

$D = R \ {- \frac{1}{2}, 0, \frac{1}{2}}$

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

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

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

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

$c)$

$za ł o \overset{z}{˙} enia:$

$⎩ ⎨ ⎧ x^{2} - 25 \neq = 0 x^{2} - 5 x \neq = 0 x^{2} + 5 x \neq = 0$

$⎩ ⎨ ⎧ (x - 5) (x + 5) \neq = 0 x (x - 5) \neq = 0 x (x + 5) \neq = 0$

$⎩ ⎨ ⎧ x \neq = 5 i x \neq = - 5 x \neq = 0 i x \neq = 5 x \neq = 0 i x \neq = - 5$

$D = R \ {- 5, 0, 5}$

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

$= \frac{x ( x + 1 )}{x ( x - 5 ) ( x + 5 )} - \frac{( x + 1 ) ( x + 5 )}{x ( x - 5 ) ( x + 5 )} + \frac{( x - 1 ) ( x - 5 )}{x ( x - 5 ) ( x + 5 )} =$

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

$= \frac{x ^{2} + x - x ^{2} - 5 x - x - 5 + x ^{2} - 5 x - x + 5}{x ( x - 5 ) ( x + 5 )} = \frac{x ^{2} - 11 x}{x ( x - 5 ) ( x + 5 )} =$

$= \frac{x ( x - 11 )}{x ( x - 5 ) ( x + 5 )} = \frac{x - 11}{( x - 5 ) ( x + 5 )}$

$d)$

$za ł o \overset{z}{˙} enia:$

$⎩ ⎨ ⎧ x^{2} - 2 x + 1 \neq = 0 x \neq = 0 x - 1 \neq = 0$

$⎩ ⎨ ⎧ (x - 1)^{2} \neq = 0 x \neq = 0 x \neq = 1$

$⎩ ⎨ ⎧ x \neq = 1 x \neq = 0 x \neq = 1$

$D = R \ {0, 1}$

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

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

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

Agnieszka Nowak

Nauczycielka matematyki

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