Question

​Clasa a X-a.
Vă rog frumos să mă ajutați. Urgent!
Vă rog frumos să mă ajutați la punctele d), e), f), g), h).
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Answer 1

Răspuns:

$x^{2}-4x+5=0=>\\x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^{2}-4\cdot1\cdot5}}{2\cdot1};\\\ x_{2}=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^{2}-4\cdot1\cdot5}}{2\cdot1}=>\\x=\frac{4+\sqrt{16-20}}{2};\ x_{2}=\frac{4-\sqrt{16-20}}{2}=>\\x=\frac{4+\sqrt{-4}}{2};\ x_{2}=\frac{4-\sqrt{-4}}{2}=>\\x=2+i,\ x_{2}=2-i$

$3z-2=\frac{1}{z}=>\\\left(3z-2\right)z-1=0=>\\3z^{2}+z-3z-1=0=>\\\left(3z+1\right)\left(z-1\right)=0=>\\3z=1=0;\ z-1=0=>\\z=-\frac{1}{3};\ z=1$

$z\left(z-1\right)^{2}=z^{3}+1=>\\z\left(z^{2}-2z+1\right)=z^{3}+1=>\\z^{3}-2z^{2}+z=z^{3}+1=>\\-2z^{2}+z=1=>\\-2z^{2}+z-1=0=>\\2z^{2}-z+1=0=>\\z=\frac{-\left(-1\right)+\sqrt{\left(-1\right)^{2}-4\cdot2\cdot1}}{2\cdot2};\ z_{2}=\frac{-\left(-1\right)-\sqrt{\left(-1\right)^{2}-4\cdot2\cdot1}}{2\cdot2}=>\\z=\frac{1+\sqrt{-7}}{4};\ z_{2}=\frac{1-\sqrt{-7}}{4}$

$\left(z-1\right)^{2}+\left(2z-1\right)^{2}=0=>\\z^{2}-2z+1+4z^{2}-4z+1=0=>\\5z^{2}-6z+2=0=>\\z=\frac{-\left(-6\right)+\sqrt{\left(-6\right)^{2}-4\cdot5\cdot2}}{2\cdot5};\ z_{2}=\frac{-\left(-6\right)-\sqrt{\left(-6\right)^{2}-4\cdot5\cdot2}}{2\cdot5}=>\\z=\frac{6+\sqrt{36-40}}{10};\ z_{2}=\frac{6-\sqrt{36-40}}{10}=>\\z=\frac{6+\sqrt{-4}}{10};\ z_{2}=\frac{6-\sqrt{-4}}{10}$

$z\left(z^{2}+z+1\right)\left(z^{2}+25\right)=0=>\\z^{2}+z+1=0;\ z_{62}+25=0=>\\z=0$