Question

Me ajudem pfvvvvv
resolva as equações irracionais
a)√(3x+1)+6=2
b)√(5x-10)=√(3x+2)
c)3√(x+1)=√18
d)X=√(6-x)
e)X-6=√x
f)√(3x+6)-2=x

Answer 1

$a) \sqrt{ (3x+1)}+6=2\\\\ ( \sqrt{ (3x+1)}+6)^2=(2)^2\\\\ 3x+1+36=4\\\\ 3x=4-1-36\\\\ 3x=-33\\\\ x= \frac{-33}{3} \\\\ x=-11$





$b) \sqrt{ (5x-10)}= \sqrt{(3x+2)}\\\\ (\sqrt{ (5x-10)})^2= (\sqrt{(3x+2)})^2\\ \\5x-10=3x+2\\\\ 5x-3x=2+10\\\\ 2x=12\\\\ x= \frac{12}{2} \\\\ x=6$





$c)3 \sqrt{ (x+1)}= \sqrt{ 18}\\\\ (3 \sqrt{ (x+1)})^2=( \sqrt{ 18})^2\\\\ 9*x+1=18\\\\ 9x+9=18\\\\ 9x=18-9\\ \\9x=9\\\\ x= \frac{9}{9} \\ \\x=1$





$d)x= \sqrt{ (6-x)}\\\\ ( x)^{2} = ( \sqrt{ (6-x)})^2\\\\ x^{2} =6-x\\\\ x^{2} +x-6=0\\\\ a=1;b=1;c=-6\\\\ \Delta = b^{2} -4ac\to \Delta = 1^2-4*1*(-6)\to \Delta = 1+24\to \Delta = 25\\ \\x= \frac{-b+- \sqrt{\Delta } }{2a} \to x= \frac{-1+- \sqrt{25 } }{2*1} \to x= \frac{-1+-5}{2} \to\\ \\ x'= \frac{-1+5}{2} \to x'= \frac{4}{2} \to x'=2\\\\ x''= \frac{-1-5}{2} \to x''= \frac{-6}{2} \to x''=-3\\\\\\ S=(-3;2)$





$e)x-6= \sqrt{x}\\\\ (x-6)^2=( \sqrt{x})^2\\\\ x^{2} -12x+36= x^{2} \\ \\x^{2} -12x+36- x^{2}=0 \\\\ -12x+36=0\\\\ -12x=-36\\\\ x= \frac{-36}{-12}\\\\x=3$





$f)\sqrt{ (3x+6)}-2=x\\\\ (\sqrt{ (3x+6)}-2)^2=( x)^{2} \\\\3x+6-4= x^{2} \\\\ x^{2} -3x-6+4=0\\\\ x^{2} -3x-2=0\\\\ a=1;b=-3;c=-2\\\\\\ \Delta=b^2-4ac\to \Delta=(-3)^2-4*1*(-2)\to \Delta=9+8\to \Delta=17\\\\\\ x= \frac{-b+- \sqrt{\Delta} }{2a} \to x= \frac{-(-3)+- \sqrt{17} }{2*1} \to x= \frac{3+-4,12 }{2} \to \\\\x'= \frac{3+4,12 }{2} \to x'= \frac{7,12 }{2} \to x'=3,56\\\\ x''= \frac{3-4,12 }{2} \to x''= \frac{-1,12 }{2} \to x''=0,56\\\\\\ S=(0,56;3,56)$