Klasa
II liceum
Przedmiot
Matematyka
Wybierz ksi膮偶k臋
Matematyka 2. Poziom rozszerzony. Po gimnazjum, Zbi贸r zada艅
  • 2.40

    Zadanie

  • 2.41

    Zadanie

  • 2.42

    Zadanie

  • 2.43

    Zadanie

  • 2.44

    Zadanie

  • 2.45

    Zadanie

Skorzystamy ze wzoru skr贸conego mno偶enia na r贸偶nic臋 kwadrat贸w:

a2b2=(ab)(a+b){a}^{{2}}-{b}^{{2}}={\left({a}-{b}\right)}{\left({a}+{b}\right)}

a)f(x)=9x281=(3x)292=(3x9)(3x+9){a}{)}\ {f{{\left({x}\right)}}}={9}{x}^{{2}}-{81}={\left({3}{x}\right)}^{{2}}-{9}^{{2}}={\left({3}{x}-{9}\right)}{\left({3}{x}+{9}\right)}

聽聽聽3x9=0聽聽+9聽聽聽聽聽聽3x+9=0聽聽9\ \ \ {3}{x}-{9}={0}\ \ {\left|+{9}\ \ \ \vee\ \ \ {3}{x}+{9}={0}\ \ \right|}-{9}

聽聽聽3x=9聽聽3聽聽聽聽聽聽聽聽聽聽聽聽3x=9聽聽3\ \ \ {3}{x}={9}\ \ {\left|{3}\ \ \ \ \ \ \ \ \ \vee\ \ \ {3}{x}=-{9}\ \ {\left|{3}\right.}\right.}

聽聽聽x=3聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽x=3\ \ \ {x}={3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vee\ \ \ {x}=-{3}

b)f(x)=12x2+4=412x2=22(12x)2={b}{)}\ {f{{\left({x}\right)}}}=-\frac{{1}}{{2}}{x}^{{2}}+{4}={4}-\frac{{1}}{{2}}{x}^{{2}}={2}^{{2}}-{\left(\sqrt{{\frac{{1}}{{2}}}}{x}\right)}^{{2}}=聽聽22(12x)2={2}^{{2}}-{\left(\frac{{1}}{\sqrt{{2}}}{x}\right)}^{{2}}=

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