Question

Poderiam me ajudar pfv?? estou precisando imediatamente, obrigado

Answer 1

4.

a) $\sqrt{(5x+1)}+1=x$

$\sqrt{(5x+1)}=x-1$

$(\sqrt{(5x+1)})^2=(x-1)^2$

$5x+1=x^2-2x+1$

$x^2-7x=0$

$x\cdot(x-7)=0$

$x'=0$

$x-7=0~\rightarrow~x"=7$

Para $x=0$:

$\sqrt{(5\cdot0+1)}+1=1$

$\sqrt{1}+1=1$

$1+1=1$ (falso)

Para $x=7$:

$\sqrt{(5\cdot7+1)}+1=7$

$\sqrt{36}+1=7$

$6+1=7$ (verdadeiro)

$S=\{7\}$

b) $\sqrt{y+10}-\sqrt{2y-5}=0$

$\sqrt{y+10}=\sqrt{2y-5}$

$(\sqrt{y+10})^2=(\sqrt{2y-5})^2$

$y+10=2y-5$

$y=15$

$S=\{15\}$

c) $\sqrt{4a+5}=a$

$(\sqrt{4a+5})^2=a^2$

$4a+5=a^2$

$a^2-4a-5=0$

$\Delta=(-4)^2-4\cdot1\cdot(-5)=16+20=36$

$a=\dfrac{-(-4)\pm\sqrt{36}}{2\cdot1}=\dfrac{4\pm6}{2}=2\pm3$

$a'=2+3~\rightarrow~a'=5$

$a"=2-3~\rightarrow~a"=-1$

Para $a=-1$:

$\sqrt{4\cdot(-1)+5}=-1$

$\sqrt{1}=-1$ (falso)

Para $a=5$:

$\sqrt{4\cdot5+5}=5$

$\sqrt{25}=5$ (verdadeiro)

$S=\{5\}$

d) $2+p+\sqrt{2p^2+3p-2}=0$

$\sqrt{2p^2+3p-2}=-(2+p)$

$(\sqrt{2p^2+3p-2})^2=[-(2+p)]^2$

$2p^2+3p-2=4+4p+p^2$

$p^2-p-6=0$

$\Delta=(-1)^2-4\cdot1\cdot(-6)=25$

$p=\dfrac{1\pm5}{2}$

$p'=\dfrac{1+5}{2}~\rightarrow~p'=3$

$p"=\dfrac{1-5}{2}~\rightarrow~p"=-2$

Para $p=3$:

$2+3+\sqrt{2\cdot3^2+3\cdot3-2}=0$

$5+\sqrt{25}=0$

$\sqrt{25}=-5$ (falso)

Para $p=-2$:

$2-2+\sqrt{2\cdot(-2)^2+3\cdot(-2)-2}=0$

$0+\sqrt{8-6-2}=0$

$\sqrt{0}=0$ (verdadeiro)

$S=\{-2\}$

5.  

a) $x^4+4x^2-21=0$

Seja $y=x^2$

$y^2+4y-21=0$

$\Delta=4^2-4\cdot1\cdot(-21)=16+84=100$

$y=\dfrac{-4\pm\sqrt{100}}{2\cdot1}=\dfrac{-4\pm10}{2}=-2\pm5$

$y'=-2+5~\rightarrow~y'=3$

$y"=-2-5~\rightarrow~y"=-7$

Para $y=-7$:

$x^2=-7$

Não há solução real.

Para $y=3$:

$x^2=3$

$x'=\sqrt{3}$

$x"=-\sqrt{3}$

$S=\{-\sqrt{3},\sqrt{3}\}$

b) $\sqrt[3]{11x+26}=5$

$(\sqrt[3]{11x+26})^3=5^3$

$11x+26=125$

$11x=99$

$x=9$

$S=\{9\}$

c) $kx^2-x-1-k=0$

$\Delta=(-1)^2-4\cdot k\cdot(-1-k)=1+4k+4k^2=(2k+1)^2$

$x=\dfrac{-(-1)\pm\sqrt{(2k+1)^2}}{2\cdot k}=\dfrac{1\pm2k+1}{2k}$

$x'=\dfrac{1+2k+1}{2k}=\dfrac{2k+2}{2k}=\dfrac{k+1}{k}$

$x"=\dfrac{1-2k-1}{2k}=\dfrac{-2k}{2k}=-1$

$S=\right\{-1,\dfrac{k+1}{k}\right\}$

d) $\sqrt{x^2-1}=2x-2$

$(\sqrt{x^2-1})^2=(2x-2)^2$

$x^2-1=4x^2-8x+4$

$3x^2-8x+5=0$

$\Delta=(-8)^2-4\cdot3\cdot5=64-60=4$

$x=\dfrac{-(-8)\pm\sqrt{4}}{2\cdot3}=\dfrac{8\pm2}{6}$

$x'=\dfrac{8+2}{6}~\rightarrow~x'=\dfrac{10}{6}~\rightarrow~x'=\dfrac{5}{3}$

$x"=\dfrac{8-2}{6}~\rightarrow~x"=1$

Para $x=1$:

$\sqrt{1^2-1}=2\cdot1-2$

$\sqrt{0}=0$ (verdadeiro)

Para $x=\dfrac{5}{3}$:

$\sqrt{\left(\dfrac{5}{3}\right)^2-1}=2\cdot\dfrac{5}{3}-2$

$\sqrt{\dfrac{25}{9}-1}=\dfrac{10}{3}-2$

$\sqrt{\dfrac{16}{9}=\dfrac{4}{3}$ (verdadeiro)

$S=\left\{1,\dfrac{5}{3}\right\}$