To solve for \(\displaystyle x\), square both terms to cancel out the square root sign on the left-hand side.

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

\(\displaystyle x-2=1\)

Next, add two to both sides of the equation.

\(\displaystyle x-2=1\)

     \(\displaystyle +2\)  \(\displaystyle +2\)

_____________

\(\displaystyle x=3\)

From here, check for extraneous solutions by substituting the value found for \(\displaystyle x\) into the original equation.

\(\displaystyle \sqrt{x-2}=1\)

\(\displaystyle \\ \sqrt{3-2}=1 \\\sqrt{1}=1 \\\pm1=1\)

Since the square root of one is either positive or negative one, the solution found is verified.

To solve for \(\displaystyle x\), first subtract one from both sides.

    \(\displaystyle 1-\sqrt{x+1}=5\)

\(\displaystyle -1\)                        \(\displaystyle -1\)

____________________

\(\displaystyle -\sqrt{x+1}=4\)

From here, divide both sides by negative one.

\(\displaystyle -\frac{\sqrt{x+1}}{-1}=\frac{4}{-1}\)

\(\displaystyle \sqrt{x+1}=-4\)

Next, square both sides to cancel the square root sign on the left-hand side.

\(\displaystyle (\sqrt{x+1})^2=(-4)^2\)

\(\displaystyle x+1=16\)

Now, subtract one from both sides to solve for \(\displaystyle x\).

\(\displaystyle x+1=16\)

     \(\displaystyle -1\)     \(\displaystyle -1\)

____________

\(\displaystyle x=15\)

Lastly, check for extraneous solutions by substituting the value found for \(\displaystyle x\) into the original equation.

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

\(\displaystyle \\1-\sqrt{15+1}=5 \\1-\sqrt{16}=5 \\1-\pm4=5\)

Thus, the answer is verified.

To solve for \(\displaystyle x\), add two to both sides of the equation.

\(\displaystyle \frac{1}{x}-2=10\)

     \(\displaystyle +2\)     \(\displaystyle +2\)

______________

\(\displaystyle \frac{1}{x}=12\)

Now, multiply by \(\displaystyle x\) on both sides. This will move the variable from the denominator on one side to the numerator of the other side.

\(\displaystyle x\cdot \frac{1}{x}=12\cdot x\)

\(\displaystyle 1=12x\)

Lastly, divide both sides by twelve.

\(\displaystyle \frac{1}{12}=\frac{12x}{12}\)

The twelve in the numerator and the twelve in the denominator cancel out thus, solving for \(\displaystyle x\).

\(\displaystyle \frac{1}{12}=x\)

To solve for \(\displaystyle x\), first subtract four from both sides so all constants are on one side.

\(\displaystyle \sqrt{x}+4=3\)

         \(\displaystyle -4\)  \(\displaystyle -4\)

_____________

\(\displaystyle \sqrt{x}=-1\)

Now, square both sides to cancel the square root sign on the left-hand side.

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

\(\displaystyle x=1\)

Recall that when a negative number is squared the result is always a positive value.

From here, check for extraneous solutions by substituting in the value found for \(\displaystyle x\).

\(\displaystyle \\\sqrt{x}+4=3 \\\sqrt{1}+4=3 \\\pm1+4=3 \\-1+4=3\)

Since the square root of a value results in a positive and a negative value, our solution is verified.

To solve for \(\displaystyle x\) start by squaring both sides to eliminate the square root sign.

\(\displaystyle \left( \sqrt{\frac{x}{2}}\right)^2=(6)^2\)

\(\displaystyle \frac{x}{2}=36\)

From here multiply both sides by two. On the left-hand side, the two in the numerator will cancel out the two in the denominator.

\(\displaystyle 2\cdot \frac{x}{2}=36\cdot 2\)

\(\displaystyle x=72\)

From here, check for extraneous solutions by substituting in the value found for \(\displaystyle x\) into the original equation.

\(\displaystyle \sqrt{\frac{x}{2}}=6\)

\(\displaystyle \sqrt{\frac{72}{2}}=6\)

\(\displaystyle \sqrt{36}=6\)

\(\displaystyle \pm6=6\)

Thus, the solution is verified.

To solve for \(\displaystyle x\), first add two to both sides.

\(\displaystyle \frac{3}{x}-2=5\)

     \(\displaystyle +2\)   \(\displaystyle +2\)

____________

\(\displaystyle \frac{3}{x}=7\)

From here, multiply both sides by \(\displaystyle x\). 

\(\displaystyle x\cdot \frac{3}{x}=7\cdot x\)

\(\displaystyle 3=7x\)

Now divide by seven to solve for \(\displaystyle x\).

\(\displaystyle \frac{3}{7}=\frac{7x}{7}\)

The seven in the numerator and seven in the denominator cancel out, thus solving for \(\displaystyle x\).

\(\displaystyle \frac{3}{7}=x\)

To solve for \(\displaystyle x\), first square both sides of the equation. Squaring a square root sign will cancel them out.

\(\displaystyle (\sqrt{5-x})^2=(2)^2\)

\(\displaystyle 5-x=4\)

Now, subtract five from both sides to get all constants on one side of the equation while all variables are on the other side of the equation.

   \(\displaystyle 5-x=4\)

\(\displaystyle -5\)           \(\displaystyle -5\)

_____________

\(\displaystyle -x=-1\)

From here, divide by negative one on both sides.

\(\displaystyle \frac{-x}{-1}=\frac{-1}{-1}\)

\(\displaystyle x=1\)

Lastly, check for extraneous solutions by substituting in the value found for \(\displaystyle x\) into the original equation.

\(\displaystyle \\\sqrt{5-1}=2 \\\sqrt{4}=2 \\\pm2=2\)

Thus, the solution is verified.

To solve for \(\displaystyle x\), simply multiply both sides of the equation by two.

\(\displaystyle \frac{x}{2}=8\)

\(\displaystyle 2\cdot \frac{x}{2}=8\cdot 2\)

The two in the numerator cancels the two in the denominator, thus solving for \(\displaystyle x\).

\(\displaystyle x=16\)

To solve for \(\displaystyle x\), first divide by two.

\(\displaystyle 2\sqrt{x}=14\)

\(\displaystyle \frac{2\sqrt{x}}{2}=\frac{14}{2}\)

The two in the denominator cancels the two in the numerator.

\(\displaystyle \sqrt{x}=7\)

From here, square both sides of the equation. Squaring a square root cancels it out.

\(\displaystyle (\sqrt{x})^2=(7)^2\)

\(\displaystyle x=49\)

To check for extraneous solutions, simply substitute into the original equation the value found for \(\displaystyle x\).

\(\displaystyle 2\sqrt{x}=14\)

\(\displaystyle \\2\sqrt{49}=14 \\2\cdot \pm 7=14 \\2(7)=14 \\2(-7)=-14\neq 14\)

To solve for \(\displaystyle x\), add three to both sides of the equation.

\(\displaystyle x^2-3=13\)

      \(\displaystyle +3\)     \(\displaystyle +3\)

______________

\(\displaystyle x^2=16\)

From here, take the square root of both sides. Taking the square root of a squared term eliminates the square.

\(\displaystyle \sqrt{x^2}=\sqrt{16}\)

\(\displaystyle x=\pm4\)

Next, check for extraneous solutions. Substitute each potential solution into the original equation to verify if it results in a legal solution.

\(\displaystyle x^2-3=13\)

Substituting in positive four.

\(\displaystyle \\4^2-3=13 \\16-3=13 \\13=13\)

Substituting in negative four.

\(\displaystyle \\(-4)^2-3=13 \\16-3=13 \\13=13\)

Therefore, both answers are verified solutions.