Question

Ajutatima cu acest exercitiu
.p.s cum e mai bine sa dau intrebarile, sa scriu si intrebarea sau sa pun numai poza ?

Answer 1

Răspuns:

x=612

Explicație pas cu pas:

C.E.

x+13≥0

x≥-13

fie√de ordin 4din (x+13)=t, t>0

atunci

t+20=t²

t²-t-20=0

cu Viete, Δ sau descompunere , obtii rapid

t1=-4..........√ordin 4 din(x+13)=-5.........x∈∅

t2=5....√ordin4 din (x+13)=5..........

ridicam la puterea 4..........x+13=625.....x=612>-13

verific

√ordin4 din625+20=√625

5+20=25

adevarat , bine rezolvat

Answer 2

 

$\displaystyle\\ \text{Rezolvam ecuatia:}\\\\\sqrt[\b4]{x+13}+20=\sqrt{x+13}\\\\\text{Folosim formula:}~~~~~\sqrt[\b4]{a}= \sqrt[\b2]{\sqrt[\b2]{a~~}}=\sqrt{\sqrt{a~~}}\\\\\sqrt{\sqrt{x+13}}+20=\sqrt{x+13}\\\\\text{Facem substitutia: }~~~\boxed{y=\sqrt{x+13}}\\\\\sqrt{y}+20=y\\\\\sqrt{y}=y-20~\Big| \text{Ridicam la puterea a 2-a.}\\y=(y-20)^2\\\\y=y^2-40y+400\\\\y^2-40y+400-y=0\\\\y^2-41y+400-y=0\\\\y_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{41\pm\sqrt{1681-4\cdot400}}{2}=$

$\displaystyle\\=\frac{41\pm\sqrt{1681-1600}}{2}=\frac{41\pm\sqrt{81}}{2}=\frac{41\pm9}{2}\\\\y_1=\frac{41-9}{2}=\frac{32}{2}=\boxed{16}\\\\y_2=\frac{41+9}{2}=\frac{50}{2}=\boxed{25}\\\\\text{Ne intoarcem la substitutia: }~~\boxed{y=\sqrt{x+13}}\\\\\sqrt{x+13}=16\implies x+13=16^2\implies x_1=256-13=\boxed{243}\\\\\sqrt{x+13}=25\implies x+13=25^2\implies x_2=625-13=\boxed{612}$

$\displaystyle\\\text{Deoarece am ridicat la puterea a 2-a pentru a scapa de radical,}\\\\\text{trebuie sa verificam solutiile, stiind ca ridicarea la putere a unei ecuatii}\\\\\text{poate sa introduca radacini false}\text{Verificare: }\\\\\text{Verificam pe }~~x_1=243\\\\\sqrt[\b4]{243+13}+20=\sqrt{243+13}\\\\\sqrt[\b4]{256}+20=\sqrt{256}\\\\4+20=16\\\\24=16~~~\text{Fals}\implies~x_1~\text{nu este solutie}$

$\displaystyle\\\text{Verificam pe }~~x_2=612\\\\\sqrt[\b4]{612+13}+20=\sqrt{612+13}\\\\\sqrt[\b4]{625}+20=\sqrt{625}\\\\5+20=25\\\\25=25~~~\text{Corect}\\\\\implies~\boxed{x=612}~\text{este solutie unica}$