Question

am nevoie de ele urgent ,va rog......​

Answer 1

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

$\displaystyle 2.~~\sqrt[3]{x^2-x-3} =-1 \Rightarrow \Big(\sqrt[3]{x^2-x-3}\Big)^3=(-1)^3 \Rightarrow x^2-x-3=-1\Rightarrow \\\\\Rightarrow x^2-x-3+1=0\Rightarrow x^2-x-2=0\\\\ \Delta=(-1)^2-4 \cdot 1 \cdot (-2)=1+8=9 > 0\\\\ x_1=\frac{-(-1)-\sqrt{9} }{2 \cdot 1} =\frac{1-3}{2} =\frac{-2}{2} =-1\\\\ x_2=\frac{-(-1)+\sqrt{9} }{2 \cdot 1} =\frac{1+3}{2} =\frac{4}{2} =2$

$\displaystyle 3.~~\sqrt{3x+4} =2\sqrt{x} ~~~~~~~~~~~~~~~~~~~~~~C.E.~3x+4\geq 0;~x\geq 0\\\\ \sqrt{3x+4} =2\sqrt{x} \Rightarrow \Big(\sqrt{3x+4}\Big)^2=\Big(2\sqrt{x} \Big)^2\Rightarrow 3x+4=4x \Rightarrow \\\\ \Rightarrow 3x-4x=-4 \Rightarrow -x=-4 \Rightarrow x=4$

$\displaystyle 4.~~\sqrt[3]{x^3+x^2-x-2} =x \Rightarrow \Big(\sqrt[3]{x^3+x^2-x-2}\Big)^3=x^3 \Rightarrow \\\\ \Rightarrow x^3+x^2-x-2=x^3 \Rightarrow x^3+x^2-x-2-x^3=0\Rightarrow x^2-x-2=0\\\\ \Delta=(-1)^2 -4\cdot1\cdot(-2)=1+8=9 > 0\\\\ x_1=\frac{-(-1)-\sqrt{9} }{2\cdot1}=\frac{1-3}{2} =\frac{-2}{2} =-1 \\\\x_2=\frac{-(-1)+\sqrt{9} }{2\cdot1}=\frac{1+3}{2} =\frac{4}{2} =2$

$\displaystyle 5.~~\sqrt{x+6} =x ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x+6\geq 0\\\\\sqrt{x+6} =x \Rightarrow \Big(\sqrt{x+6}\Big)^2=x^2\Rightarrow x+6=x^2 \Rightarrow -x^2+x+6=0\Big|\cdot(-1)\Rightarrow \\\\ \Rightarrow x^2-x-6=0\\\\ \Delta=(-1)^2-4 \cdot 1 \cdot (-6)=1+24=25 > 0\\\\ x_1=\frac{-(-1)-\sqrt{25} }{2\cdot1}=\frac{1-5}{2} =\frac{-4}{2} =-2 \not \in C.E.\\\\ x_2=\frac{-(-1)-\sqrt{25} }{2\cdot1}=\frac{1+5}{2} =\frac{6}{2} =3$

$\displaystyle 6.~~\sqrt{x^2-x-2} =x-2~~~~~~~~~~~~~~~~~~~~~~x^2-x-2\geq 0\\\\ \sqrt{x^2-x-2} =x-2 \Rightarrow \Big(\sqrt{x^2-x-2} \Big)^2=(x-2)^2\Rightarrow \\\\ \Rightarrow x^2-x-2=x^2-2 \cdot x \cdot 2+2^2 \Rightarrow x^2-x-2=x^2-4x+4 \Rightarrow \\\\ \Rightarrow x^2-x^2-x+4x=4+2 \Rightarrow 3x=6 \Rightarrow x=\frac{6}{3} \Rightarrow x=2$

$\displaystyle 7.~~\sqrt{x^2-3} -x=-1~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x^2-3\geq 0\\\\ \sqrt{x^2-3} -x=-1 \Rightarrow \sqrt{x^2-3} =x-1 \Rightarrow \Big(\sqrt{x^2-3}\Big)^2=(x-1)^2\Rightarrow \\\\ \Rightarrow x^2-3=x^2-2\cdot x \cdot1+1^2 \Rightarrow x^2-3=x^2-2x+1 \Rightarrow \\\\ \Rightarrow x^2-x^2+2x=1+3 \Rightarrow 2x=4 \Rightarrow x=\frac{4}{2} \Rightarrow x=2$

$\displaystyle 8.~~\sqrt{25-x^2} =4 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~25-x^2\geq 0\\\\ \sqrt{25-x^2} =4\Rightarrow \Big(\sqrt{25-x^2} \Big)^2=4^2 \Rightarrow 25-x^2=16 \Rightarrow -x^2=16-25 \Rightarrow \\\\ \Rightarrow -x^2=-9 \Rightarrow x^2=9 \Rightarrow x=\pm \sqrt{9} \Rightarrow x=\pm 3$

$\displaystyle 9.~~\sqrt{x^2+x-4} =2\sqrt{2} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x^2+x-4\geq 0\\\\ \sqrt{x^2+x-4} =2\sqrt{2} \Rightarrow \Big(\sqrt{x^2+x-4}\Big)^2 =\Big(2\sqrt{2} \Big)^2\Rightarrow \\\\ \Rightarrow x^2+x-4=8 \Rightarrow x^2+x-4-8=0 \Rightarrow x^2+x-12=0\\\\ \Delta=1^2-4 \cdot 1 \cdot (-12)=1+48=49 > 0\\\\ x_1=\frac{-1-\sqrt{49} }{2\cdot1}=\frac{-1-7}{2} =\frac{-8}{2} =-4\\\\ x_2=\frac{-1+\sqrt{49} }{2\cdot1}=\frac{-1+7}{2} =\frac{6}{2} =3$

$\displaystyle 10.~~\sqrt{x+1} =3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x+1\geq 0\\\\ \sqrt{x+1} =3 \Rightarrow \Big(\sqrt{x+1} \Big)^2=3^2 \Rightarrow x+1=9 \Rightarrow x=9-1 \Rightarrow x=8$