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Trebuie rezolvate cu delta va rog mult puncte maxime și coronița
Nu va bateți joc!


Trebuie Rezolvate Cu Delta Va Rog Mult Puncte Maxime Și Coronița Nu Va Bateți Joc class=

Răspuns :

1)

[tex]\it x^2-3x+2=0,\ a=1,\ b=-3,\ c=2\\ \\ \Delta=b^2-4ac=(-3)^2-4\cdot1\cdot2=9-8=1\\ \\ \\ x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{3\pm1}{2} \Rightarrow\begin{cases}\ \it x_1=\dfrac{3-1}{2}=\dfrac{2}{2}=1\\ \\ \it x_2=\dfrac{3+1}{2}=\dfrac{4}{2}=2\end{cases}[/tex]

3)

[tex]\it x^2-2x+1=0 \Rightarrow (x-1)^2=0 \Rightarrow x_1=x_2=1[/tex]

4)

[tex]\it-x^2+x-7=0|_{\cdot(-1)} \Rightarrow x^2-x+7=0,\ a=1,\ b=-1,\ c=7\\ \\ \Delta =b^2-4ac=(-1)^2-4\cdot1\cdot7=1-28=-27<0 \Rightarrow x_{1,2}\not\in\mathbb{R}[/tex]

5)

[tex]\it 2x^2-8x+1=0,\ a=2,\ b=-8,\ c=1\\ \\ \Delta=b^2-4ac=(-8)^2 -4\cdot2\cdot1=64-8=56\\ \\ x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{8\pm\sqrt{56}}{2\cdot2}=\dfrac{8\pm\sqrt{4\cdot14}}{4}=\dfrac{8\pm2\sqrt{14}}{4}=\dfrac{2(4\pm\sqrt{14})^{(2}}{4}=\\ \\ \\ =\dfrac{4\pm\sqrt{14}}{2}\ \begin{cases} \it x_1=\dfrac{4-\sqrt{14}}{2}\\ \\ \\ \it x_2=\dfrac{4+\sqrt{14}}{2}\end{cases}[/tex]