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Matematyka 2. Zakres rozszerzony. Reforma 2019, Zbiór zadań
  • 3.139

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  • 3.141

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    Zadanie

  • 3.146

    Zadanie

a(x24)2=9(x24){a}\text{)}\ {\left({x}^{{2}}-{4}\right)}^{{2}}={9}{\left({x}^{{2}}-{4}\right)} 

(x24)29(x24)=0{\left({x}^{{2}}-{4}\right)}^{{2}}-{9}{\left({x}^{{2}}-{4}\right)}={0} 

(x24)(x249)=0{\left({x}^{{2}}-{4}\right)}{\left({x}^{{2}}-{4}-{9}\right)}={0} 

(x24)(x213)=0{\left({x}^{{2}}-{4}\right)}{\left({x}^{{2}}-{13}\right)}={0} 

zatem:

x24=0    x213=0{x}^{{2}}-{4}={0}\ \ \vee\ \ {x}^{{2}}-{13}={0} 

x2=4    x2=13{x}^{{2}}={4}\ \ \vee\ \ {x}^{{2}}={13} 

x=2    x=2    x=13    x=13{x}=-{2}\ \ \vee\ \ {x}={2}\ \ \vee\ \ {x}=-\sqrt{{13}}\ \ \vee\ \ {x}=\sqrt{{13}} 

więc:

x{13,2,2,13}{x}\in{\left\lbrace-\sqrt{{13}},-{2},{2},\sqrt{{13}}\right\rbrace} 


b(x2+3)25(x2+3)+4=0{b}\text{)}\ {\left({x}^{{2}}+{3}\right)}^{{2}}-{5}{\left({x}^{{2}}+{3}\right)}+{4}={0} 

Stosujemy podstawienie:

t=x2+3, t3{t}={x}^{{2}}+{3},\ {t}\ge{3} 

t25t+4=0{t}^{{2}}-{5}{t}+{4}={0} 

Δ=(5)2414=2516=9, Δ=3\Delta={\left(-{5}\right)}^{{2}}-{4}\cdot{1}\cdot{4}={25}-{16}={9},\ \sqrt{\Delta}={3} 

t1=532=22=1<3{t}_{{1}}=\frac{{{5}-{3}}}{{2}}=\frac{{2}}{{2}}={1}<{3} 

t2=5+32=82=4>3{t}_{{2}}=\frac{{{5}+{3}}}{{2}}=\frac{{8}}{{2}}={4}>{3} 

zatem:

x2+3=4{x}^{{2}}+{3}={4} 

x2=1{x}^{{2}}={1} 

x=1    x=1{x}={1}\ \ \vee\ \ {x}=-{1} 

więc:

x{1,1}{x}\in{\left\lbrace-{1},{1}\right\rbrace} 


c(x2+2x)2+9=6x2+12x{c}\text{)}\ {\left({x}^{{2}}+{2}{x}\right)}^{{2}}+{9}={6}{x}^{{2}}+{12}{x} 

(x2+2x)2+9=6(x2+2x){\left({x}^{{2}}+{2}{x}\right)}^{{2}}+{9}={6}{\left({x}^{{2}}+{2}{x}\right)} 

Stosujemy podstawienie:

x2+2x=t, t(,20,+){x}^{{2}}+{2}{x}={t},\ {t}\in{\left(-\infty,-{2}\right\rangle}\cup{\left\langle{0},+\infty\right)} 

t2+9=6t{t}^{{2}}+{9}={6}{t} 

t26t+9=0{t}^{{2}}-{6}{t}+{9}={0} 

Δ=(6)2419=3636=0\Delta={\left(-{6}\right)}^{{2}}-{4}\cdot{1}\cdot{9}={36}-{36}={0} 

t0=62=3{t}_{{0}}=\frac{{6}}{{2}}={3} 

zatem:

x2+2x=3{x}^{{2}}+{2}{x}={3} 

x2+2x3=0{x}^{{2}}+{2}{x}-{3}={0} 

Δ=44(3)=4+12=16,Δ=4\Delta={4}-{4}\cdot{\left(-{3}\right)}={4}+{12}={16},\sqrt{\Delta}={4} 

x1=242=62=3{x}_{{1}}=\frac{{-{2}-{4}}}{{2}}=\frac{{-{6}}}{{2}}=-{3} 

x2=2+42=22=1{x}_{{2}}=\frac{{-{2}+{4}}}{{2}}=\frac{{2}}{{2}}={1} 

więc:

x{3,1}{x}\in{\left\lbrace-{3},{1}\right\rbrace} 


d(x28x)2+5=2(x28x){d}\text{)}\ {\left({x}^{{2}}-{8}{x}\right)}^{{2}}+{5}={2}{\left({x}^{{2}}-{8}{x}\right)} 

Stosujemy podstawienie:

x28x=t,t(,08,+){x}^{{2}}-{8}{x}={t},{t}\in{\left(-\infty,{0}\right\rangle}\cup{\left\langle{8},+\infty\right)} 

t2+5=2t{t}^{{2}}+{5}={2}{t} 

t22t+5=0{t}^{{2}}-{2}{t}+{5}={0} 

Δ=(2)2415=420<0\Delta={\left(-{2}\right)}^{{2}}-{4}\cdot{1}\cdot{5}={4}-{20}<{0} 

To równanie nie ma rozwiązania.


 e(x2x)2=20x220x{e}\text{)}\ {\left({x}^{{2}}-{x}\right)}^{{2}}={20}{x}^{{2}}-{20}{x}  

(x2x)2=20(x2x){\left({x}^{{2}}-{x}\right)}^{{2}}={20}{\left({x}^{{2}}-{x}\right)} 

(x2x)220(x2x)=0{\left({x}^{{2}}-{x}\right)}^{{2}}-{20}{\left({x}^{{2}}-{x}\right)}={0} 

(x2x)(x2x20)=0{\left({x}^{{2}}-{x}\right)}{\left({x}^{{2}}-{x}-{20}\right)}={0} 

zatem:

1x2x=0    2x2x20=0{1}\text{)}\ {x}^{{2}}-{x}={0}\ \ \vee\ \ {2}\text{)}\ {x}^{{2}}-{x}-{20}={0} 

1x(x1)=0{1}\text{)}\ {x}{\left({x}-{1}\right)}={0} 

x=0    x1=0{x}={0}\ \ \vee\ \ {x}-{1}={0} 

x=0    x=1{x}={0}\ \ \vee\ \ {x}={1} 

2x2x20=0{2}\text{)}\ {x}^{{2}}-{x}-{20}={0} 

Δ=(1)241(20)=1+80=81, Δ=9\Delta={\left(-{1}\right)}^{{2}}-{4}\cdot{1}\cdot{\left(-{20}\right)}={1}+{80}={81},\ \sqrt{\Delta}={9} 

x1=192=82=4{x}_{{1}}=\frac{{{1}-{9}}}{{2}}=\frac{{-{8}}}{{2}}=-{4} 

x2=1+92=102=5{x}_{{2}}=\frac{{{1}+{9}}}{{2}}=\frac{{10}}{{2}}={5} 

więc:

x{4,0,1,5}{x}\in{\left\lbrace-{4},{0},{1},{5}\right\rbrace} 


f(x2+3x1)2+7(x2+3x1)=12{f}\text{)}\ {\left({x}^{{2}}+{3}{x}-{1}\right)}^{{2}}+{7}{\left({x}^{{2}}+{3}{x}-{1}\right)}=-{12} 

Stosujemy podstawienie:

t=x2+3x1{t}={x}^{{2}}+{3}{x}-{1} 

t2+7t=12{t}^{{2}}+{7}{t}=-{12} 

t2+7t+12=0{t}^{{2}}+{7}{t}+{12}={0} 

Δ=724112=4948=1\Delta={7}^{{2}}-{4}\cdot{1}\cdot{12}={49}-{48}={1} 

t1=712=82=4{t}_{{1}}=\frac{{-{7}-{1}}}{{2}}=\frac{{-{8}}}{{2}}=-{4} 

t2=7+12=62=3{t}_{{2}}=\frac{{-{7}+{1}}}{{2}}=\frac{{-{6}}}{{2}}=-{3} 

zatem:

14=x2+3x1    23=x2+3x1{1}\text{)}\ -{4}={x}^{{2}}+{3}{x}-{1}\ \ \vee\ \ {2}\text{)}\ -{3}={x}^{{2}}+{3}{x}-{1} 

1x2+3x+3=0    2x2+3x+2=0{1}\text{)}\ {x}^{{2}}+{3}{x}+{3}={0}\ \ \vee\ \ {2}\text{)}\ {x}^{{2}}+{3}{x}+{2}={0} 

1Δ=912<0{1}\text{)}\ \Delta={9}-{12}<{0} 

czyli to równanie nie ma rozwiązania

2Δ=98=1, Δ=1{2}\text{)}\ \Delta={9}-{8}={1},\ \sqrt{\Delta}={1} 

x1=312=42=2{x}_{{1}}=\frac{{-{3}-{1}}}{{2}}=\frac{{-{4}}}{{2}}=-{2} 

x2=3+12=22=1{x}_{{2}}=\frac{{-{3}+{1}}}{{2}}=\frac{{-{2}}}{{2}}=-{1} 

więc:

x{2,1}{x}\in{\left\lbrace-{2},-{1}\right\rbrace} 

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