Răspuns: Ai rezolvarea mai jos
Explicație pas cu pas:
Salutare!
[tex]\bf a)~ \dfrac{4}{x+2},~pentru ~x\in \{0,2\}[/tex]
[tex]\bf Daca ~x = 0 \implies \dfrac{4}{0+2}=\dfrac{4}{2}=\dfrac{\not4}{\not2}=\boxed{\boxed{\bf 2}}[/tex]
[tex]\it~~~~~[/tex]
[tex]\bf Daca ~x = 2 \implies \dfrac{4}{2+2}=\dfrac{4}{4}=\boxed{\boxed{\bf 1}}[/tex]
[tex]\it~~~[/tex]
[tex]\bf b)~ \dfrac{x+1}{x^{2}+1},~pentru ~x\in \{-1, \sqrt{2},\sqrt{3}-1\}[/tex]
[tex]\it~~[/tex]
[tex]\bf Daca ~x = -1 \implies \dfrac{-1+1}{(-1)^{2}+1}= \dfrac{0}{1+1}=\boxed{\boxed{\bf \dfrac{0}{2}}}[/tex]
[tex]\it~~\\[/tex]
[tex]\bf Daca ~x = \sqrt{2} \implies \dfrac{\sqrt{2} +1}{(\sqrt{2} )^{2}+1}= \dfrac{\sqrt{2} +1}{2+1}=\boxed{\boxed{\bf \dfrac{\sqrt{2} +1}{3}}}[/tex][tex]\it~~~[/tex]
[tex]\bf Daca ~x = \sqrt{3}-1 \implies \dfrac{\sqrt{3}-1+1}{(\sqrt{3}-1)^{2}+1}=\dfrac{\sqrt{3}}{(\sqrt{3})^{2}-2\cdot\sqrt{3}+1+1}=\dfrac{\sqrt{3}}{3-2\sqrt{3}+2}[/tex]
[tex]\it~~~[/tex]
[tex]\bf \dfrac{\sqrt{3}}{5-2\sqrt{3}}=\dfrac{\sqrt{3}}{5-2\sqrt{3}}\cdot\dfrac{5+2\sqrt{3}}{5+2\sqrt{3}}=\dfrac{\sqrt{3} (5+2\sqrt{3})}{(5-2\sqrt{3})(5+2\sqrt{3})}=[/tex]
[tex]\it~~[/tex]
[tex]\bf \dfrac{5\sqrt{3} +6}{5^{2}-(2\sqrt{3})^{2}}=\dfrac{5\sqrt{3} +6}{25- 4\cdot3} =\dfrac{5\sqrt{3} +6}{25-12}=\boxed{\boxed{\bf\dfrac{5\sqrt{3} +6}{13}}}[/tex]
[tex]\it~~~[/tex]
[tex]\color{red}\boxed{\boxed{\bf Formule~folosite: (a+b)(a-b)=a^{2}-b^{2}~si~(a\pm b)^{2}=a^{2} \pm2ab+b^{2}}}[/tex]
PS: Daca esti pe telefon, te rog sa glisezi spre dreapta pentru a vedea rezolvarea completa
Bafta !
#copaceibrainly
