Question

Va rog ajutati ma la tema​

Answer 1

Răspuns:

Explicație pas cu pas:

$6.)\\a.) x^{2} =0=> x=0\\b.) x^{2} =-2=> x=$∅

$c.) x^{2} =16=> x=$±$\sqrt{16}$

=>$x=4\\sau\\x=-4$

$d.) 4x^{2} =64=> x^{2} =16=> x=4 \\sau x=-4$

$e.) 3x^{2} =108=> x^{2} =36=> x=\sqrt{36}=6\\\sau \\x=-\sqrt{36}=-6$

$f.) x^{2} +4=5=> x^{2} =1=> x=\sqrt{1}=1\\sau \\x=-\sqrt{1} =-1$

$g.) 3x^{2} +2=1^{2}+(-1)^{2} => \\3x^{2} =1+1-2=>\\\3x^{2} =0=> x^{2} =0=> x=0$

$h.) x^{2} +1=(\sqrt{3})^{2} =>\\x^{2} =2=> x=\sqrt{2}\\sau \\x=-\sqrt{2}$

$i.) x^{2} =\frac{4}{25}\\=> \\x=\sqrt{\frac{4}{25} } =\frac{\sqrt{4} }{\sqrt{25} } =\frac{2}{5} \\sau \\x=-\sqrt{\frac{4}{25} }=-\frac{\sqrt{4} }{\sqrt{25} } =-\frac{2}{5}$