Liceum, 2 klasa - wielomiany.Mamy 17 przykładów, od każdego daje 5 punktów, rozwiązania mają być na .... Question from @Youper

November 2018 1 21 Report

Liceum, 2 klasa - wielomiany.

Mamy 17 przykładów, od każdego daje 5 punktów, rozwiązania mają być na kartce, zeskanowane, bądź fotka.

Daje "NAJ" pierwszej osobie, która poprawnie wykona wszystkie przykłady

krysta $1)\\x^{3}-2x^{2}-x+2=0\\ \\x^2(x-2)-(x-2)=0\\ \\(x-2)(x^{2}-1)=0\\ \\(x-2)(x-1)(x+1)=0\\ \\x-2=0\vee x-1=0\vee x+1=0\\ \\x=2\vee x=1 \vee x=-1$

$2)\\x^{3}+3x^{2}-4x-12=0\\ \\x^2(x+3)-4(x+3)=0\\ \\(x+3)(x^{2}-4)=0\\ \\(x+3)(x-2)(x+2)=0\\ \\x+3=0\vee x-2=0\vee x+2=0\\x=-3\vee x=2 \vee x=-2$

$3)\\3x^{3}-x^{2}+6x-2=0\\ \\x^2(3x-1)+2(3x-1)=0\\ \\(3x+1)(x^{2}+2)=0$

$x^2+2>0 \\ \\3x+1=0\\ \\3x=-1/:3\\ \\x=-\frac{1}{3}$

$4)\\ 5x^{3}-x^{2}-10x+2=0\\ \\x^{2}(5x-1)-2(5x-1)=0\\ \\(5x-1)(x^{2}-2)=0\\ \\(5x-1)(x-1)(x+1)=0\\ \\5x-1=0\vee x-1=0 \vee x+1=0$

$5x=1/:5\veex=1 \vee x=1\\ \\x=\frac{1}{5}\veex=1 \vee x=-1$

$5)\\ 3x^{4}-15x^{3}-6x^{2}+30x=0\\ \\3x^{3}(x-5)-6x(x-5)=0\\ \\(x-5)(3x^{3}-6x)=0\\ \\(x-5)*3x(x^{2}-2)=0\\ \\(x-5)*3x*(x-\sqrt{2})(x+\sqrt{2})=0$

$x-5=0\vee 3x=0 \vee x-\sqrt{2}=0 \vee x +\sqrt{2}=0\\ \\x=5\vee x=0 \vee x=\sqrt{2} \vee x=-\sqrt{2}$

$6)\\ x^{5}-x^{3}-x^{2}+1=0\\ \\x^{3}(x^{2}-1)-(x^{2}-1)=0\\ \\(x^{2}-1)(x^{3}-1)=0\\ \\(x-1)(x+1)(x-1)(x^{2}+x+1)=0\\ \\(x-1)(x+1)(x-1)(x^{2}+x+1)=0$

$x^{2}+x+1\\ \\\Delta =b^{2}-4ac =1-4=-3 \\ \\\Delta < 0$

nie ma pierwiastków

$x-1=0\vee x+1=0\vee x-1=0\\ \\x=1\vee x=-1 \vee x=1$

rozwiązaniem jest :

$x=1\vee x=-1$

$7)\\ x^{6}-x^{4}-x^{2}+1=0\\ \\x^{4}(x^{2}-1)-(x^{2}-1)=0\\ \\(x^{2}-1)(x^{4}-1)=0\\ \\(x-1(x+1)(x^2-1)(x^2+1)=0$

$(x-1)(x+1)(x-1)(x+1)(x^2+1)=0\\ x^2+1 >0\\ \\x-1=0\vee x+1=0\\ \\x=1\vee x= -1$

$8)\\ x^{3}+2x^{2}-x-2=0\\ \\x^2(x+2)-(x+2)=0\\ \\(x+2)(x^2-1)=0\\ \\(x+2)(x-1)(x+1)=0$

$x+2=0 \vee x-1=0 \vee x+1 = 0\\ \\x=-2\vee x=1\vee x=-1$

$9)\\ x^{3}+5x^{2}+3x+15=0\\ \\x^2(x+5)+3(x+5)=0\\ \\(x+2)(x^2+3)=0$

$x^2+3> 0\\ \\x+2=0\\ \\x=-2$

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$11)\\ x^{4}-3x^{3}=3-x\\ \\ x^{4}-3x^{3}+x-3=0\\ \\x^3(x-3)+(x-3)=0\\ \\(x-3)(x+1)(x^{2}-x+1)=0\\ \\\Delta =b^{2}-4ac =(-1)^2 - 4*1*1 =1-4=-3\\ \\\Delta <0$

nie ma pierwiastków

$12)\\ x^{4}-3x^{3}=3-x\\ \\ x^{4}-3x^{3}+x-3=0\\ \\x^3(x-3)+(x-3)=0$

$(x-3)(x+1)(x^{2}-x+1)=0\\ \\x-3=0\vee x+1=0 \\ \\x=3 \vee x=-1$

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$13)\\12x^{3}-2=3x^{2}-8x\\ \\ 12x^{3}-3x^{2}+8x-2=0\\ \\3x^2(4x-1)+2(4x-1)=0\\ \\(4x-1)(3x^2+2)=0\\ \\3x^{2}+2>0$

$4x-1=0\\ \\4x=1/:4\\ \\x=\frac{1}{4}$

$14)\\5x^{3}+6x^2+x =0\\ \\ x(5x^2+6x+1)=0\\ \\x=0 \vee 5x^2+6x+1=0\\ \\\Delta =b^{2}-4ac=6^{2}-4*5*1=36-20=16\\ \\\sqrt{\Delta }=\sqrt{16}=4$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-6-4}{2*5}=\frac{-10}{10}=-1\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-6+4}{2*5}=\frac{-2}{10}=-\frac{2}{10}=-\frac{1}{5}$

rozwiazaniem jest :

$x=0 \vee x=-1\vee x=-\frac{1}{5}$

$15)\\-2x^{3}+3x^2-x =0\\ \\ -x(2x^2-3x+1)=0\\ \\-x=0 \vee 2x^2-3x+1=0\\ \\\Delta =b^{2}-4ac=(-3)^{2}-4*2*1=9-8=1\\ \\\sqrt{\Delta }=\sqrt{1}=1$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{3-1}{2*2}=\frac{2}{4}=\frac{1}{2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{3+1}{2*2}=\frac{4}{4}=1$

rozwiazaniem jest :

$x=0\vee x=1\vee x=\frac{1}{2}$

$16)\\4x^{3}+3x^2-x =0\\ \\ x(4x^2+3x-1)=0\\ \\x=0 \vee 4x^2+3x-1=0\\ \\\Delta =b^{2}-4ac=3^{2}-4*4*(-1)=9+16=25\\ \\\sqrt{\Delta }=\sqrt{25}=5$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-3-5}{2*4}=\frac{-8}{8}=-1\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-3+5}{2*4}=\frac{2}{8}=\frac{1}{4}$

rozwiazaniem jest :

$x=0\vee x=-1\vee x=\frac{1}{4}$

$17)\\6x^{3}-x^2-x =0\\ \\ x(6x^2-x-1)=0\\ \\x=0 \vee 6x^2-x-1=0\\ \\\Delta =b^{2}-4ac=(-1)^{2}-4*6*(-1)=1+24=25\\ \\\sqrt{\Delta }=\sqrt{25}=5$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{1-5}{2*6}=\frac{-4}{12}=-\frac{1}{3}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{1+5}{2*6}=\frac{6}{12}=\frac{1}{2}$

rozwiazaniem jest :

$x=0\vee x=-\frac{1}{3}\vee x=\frac{1}{2}$

1 votes Thanks 2

Youper dziękówka serdeczna :)

krysta :)