Question

Va rog frumos ajutati.ma . Dau si coroana ​

Answer 1

$\displaystyle 1.~~\sqrt{4x-3} =2~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~4x-3\geq 0\Rightarrow x\in\Big[\frac{3}{4},\infty\Big) \\\\ \sqrt{4x-3}=2 \Rightarrow \Big(\sqrt{4x-3}\Big)^2=2^2\Rightarrow 4x-3=4 \Rightarrow 4x=4+3 \Rightarrow \\\\ \Rightarrow 4x=7 \Rightarrow x=\frac{7}{4}$

$\displaystyle 2.~~\sqrt{5x+3}=6~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~5x+3\geq 0\Rightarrow x\in\Big[-\frac{3}{5},\infty\Big) \\\\\sqrt{5x+3} =6 \Rightarrow \Big(\sqrt{5x+3}\Big)^2=6^2 \Rightarrow 5x+3=36 \Rightarrow 5x=36-3\Rightarrow \\\\\Rightarrow 5x=33\Rightarrow x=\frac{33}{5}$

$\displaystyle 3.~~\sqrt{x^2+3x+5} =3~~~~~~~~~~~~~~~~~~~~~~~C.E.~x^2+3x+5\geq 0 \Rightarrow x\in(-\infty,\infty)\\\\ \sqrt{x^2+3x+5} =3 \Rightarrow \Big(\sqrt{x^2+3x+5} \Big)^2=3^2 \Rightarrow x^2+3x+5=9 \Rightarrow \\\\\Rightarrow x^2+3x+5-9=0\Rightarrow x^2+3x-4=0\\\\ \Delta=3^2-4 \cdot 1 \cdot (-4)=9+16=25 > 0\\\\ x_1=\frac{-3-\sqrt{25} }{2 \cdot 1}=\frac{-3-5}{2}=\frac{-8}{2} =-4\\\\ x_2=\frac{-3+\sqrt{25} }{2 \cdot 1}=\frac{-3+5}{2}=\frac{2}{2}=1$

$\displaystyle 4.~~\sqrt[3]{3x-2}=-2 \Rightarrow \Big(\sqrt[3]{3x-2} \Big)^3=(-2)^3\Rightarrow 3x-2=-8\Rightarrow \\\\ \Rightarrow 3x=-8+2 \Rightarrow 3x=-6 \Rightarrow x=-\frac{6}{3} \Rightarrow x=-2$

$\displaystyle 5.~~\sqrt[3]{2x-7}=4 \Rightarrow \Big(\sqrt[3]{2x-7}\Big)^3=4^3 \Rightarrow 2x-7=64 \Rightarrow 2x=64+7 \Rightarrow \\\\ \Rightarrow 2x=71 \Rightarrow x=\frac{71}{2}$

$\displaystyle 6.~~\sqrt{2x-1}=7 ~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~2x-1\geq 0 \Rightarrow x\in\Big[\frac{1}{2},\infty\Big)\\\\ \sqrt{2x-1}=7\Rightarrow \Big(\sqrt{2x-1}\Big)^2=7^2 \Rightarrow 2x-1=49 \Rightarrow 2x=49+1 \Rightarrow \\\\ \Rightarrow 2x=50 \Rightarrow x=\frac{50}{2}\Rightarrow x=25$

$\displaystyle 7.~~\sqrt{x^2-2x+10} =3~~~~~~~~~~~~~~~~~~~~C.E.~x^2-2x+10\geq 0 \Rightarrow x\in (-\infty,\infty)\\\\ \sqrt{x^2-2x+10}=3 \Rightarrow \Big(\sqrt{x^2-2x+10}\Big)^2=3^2 \Rightarrow x^2-2x+10=9\Rightarrow \\\\\Rightarrow x^2-2x+10-9=0\Rightarrow x^2-2x+1=0\\\\ \Delta=(-2)^2-4\cdot 1 \cdot 1=4-4=0\\\\ x_1=x_2=-\frac{-2}{2 \cdot 1} =1$

$\displaystyle 8.~~\sqrt{x^2-6x-18}=5 ~~~~~~~~~~~~~~~~~~C.E.~x^2-6x-18\geq 0\Rightarrow \\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x\in\Big(-\infty,-3\sqrt{3}+3 \Big] \cup\Big[3\sqrt{3}+3, \infty\Big)\\\\ \sqrt{x^2-6x-18}=5\Rightarrow \Big(\sqrt{x^2-6x-18}\Big)^2=5^2 \Rightarrow x^2-6x-18=25\Rightarrow \\\\ \Rightarrow x^2-6x-18-25=0\Rightarrow x^2-6x-43=0\\\\\Delta=(-6)^2-4 \cdot 1\cdot (-43)=36+172=208 > 0$

$\displaystyle x_1=\frac{-(-6)-\sqrt{208} }{2 \cdot 1}=\frac{6-4\sqrt{13} }{2} =\frac{2\Big(3-2\sqrt{13}\Big) }{2} =3-2\sqrt{13}\\\\ x_2=\frac{-(-6)+\sqrt{208} }{2 \cdot 1}=\frac{6+4\sqrt{13} }{2} =\frac{2\Big(3+2\sqrt{13}\Big) }{2} =3+2\sqrt{13}$

$\displaystyle 9.~~\sqrt[3]{4x+5}=3 \Rightarrow \Big(\sqrt[3]{4x+5}\Big)^3=3^3\Rightarrow 4x+5=27\Rightarrow 4x=27-5\Rightarrow\\\\\Rightarrow 4x=22\Rightarrow x=\frac{22}{4}\Rightarrow x=\frac{11}{2}$

$\displaystyle 10.~~\sqrt[3]{7x+2}=-1 \Rightarrow \Big(\sqrt[3]{7x+2}\Big)^3=(-1)^3 \Rightarrow 7x+2=-1 \Rightarrow \\\\\Rightarrow 7x=-1-2 \Rightarrow 7x=-3 \Rightarrow x=-\frac{3}{7}$

$\displaystyle 11.~~\sqrt{x^2+3x}=x ~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x^2+3x\geq 0\Rightarrow (-\infty,-3] \cup [0,\infty)\\\\ \sqrt{x^2+3x}=x \Rightarrow \Big(\sqrt{x^2+3x}\Big)^2=x^2 \Rightarrow x^2+3x=x^2 \Rightarrow x_2-x^2+3x=0\Rightarrow \\\\\Rightarrow 3x=0 \Rightarrow x=\frac{0}{3} \Rightarrow x=0$

$12.~~\sqrt{\sqrt{x+3} }=2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x+3\geq 0\Rightarrow x\in [-3,\infty)\\\\\sqrt{\sqrt{x+3} }=2 \Rightarrow \Big(\sqrt{\sqrt{x+3} }\Big)^2=2^2\Rightarrow \sqrt{x+3} =4 \Rightarrow \Big(\sqrt{x+3}\Big)^2=4^2\Rightarrow \\\\ \Rightarrow x+3=16\Rightarrow x=16-3 \Rightarrow x= 13$