Square of sum: Using multiple rules

To solve this problem, we'll follow these steps:

• Step 1: Expand $(x+y)^2$
• Step 2: Expand $(x-y)^2$
• Step 3: Rearrange and simplify the entire expression

Now, let's work through each step:
Step 1: We expand $(x+y)^2$ using the formula for the square of a sum:
$(x+y)^2 = x^2 + 2xy + y^2$.

Step 2: We expand $(x-y)^2$ using the formula for the square of a difference:
$(x-y)^2 = x^2 - 2xy + y^2$.

Step 3: Substitute these expansions back into the original expression: $(x+y)^2-(x-y)^2+(x-y)(x+y)$ becomes: $(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) + (x-y)(x+y).$

First, simplify $(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)$:
– $x^2 + 2xy + y^2 - x^2 + 2xy - y^2 = 4xy.$

Next, consider $(x-y)(x+y)$:
By using the identity for difference of squares: $(x-y)(x+y) = x^2 - y^2$.

Thus, combining our results gives:
$4xy + x^2 - y^2 = x^2 + 4xy - y^2.$

Therefore, the solution to the problem is $x^2 + 4xy - y^2$.

To solve this problem, we'll follow these steps:

• Step 1: Expand $(x+3)^2$.
• Step 2: Expand $(x-3)^2$.
• Step 3: Simplify the expression by combining like terms.

Now, let's work through each step:
Step 1: Expand $(x+3)^2$ using the formula for the square of a sum:

$(x+3)^2 = x^2 + 2 \cdot x \cdot 3 + 3^2 = x^2 + 6x + 9$

Step 2: Expand $(x-3)^2$ using the formula for the square of a difference:

$(x-3)^2 = x^2 - 2 \cdot x \cdot 3 + 3^2 = x^2 - 6x + 9$

Step 3: Add the expanded expressions together and simplify:

$(x+3)^2 + (x-3)^2 = (x^2 + 6x + 9) + (x^2 - 6x + 9)$

$(x^2 + 6x + 9) + (x^2 - 6x + 9) = 2x^2 + 0x + 18 = 2x^2 + 18$

Therefore, the solution to the problem is $2x^2 + 18$.

To solve this problem, we'll follow these steps:

• Step 1: Expand both sides of the given equation.
• Step 2: Compare and simplify the resulting expressions.
• Step 3: Solve for $a$ and $b$.

Now, let's work through each step:
Step 1: The given equation is $(a+b)(a-b) = (a+b)^2$. Let's expand both sides:
- Left side: $(a+b)(a-b) = a^2 - b^2$ based on the difference of squares formula.
- Right side: $(a+b)^2 = a^2 + 2ab + b^2$ using the square of a sum formula.

Step 2: Setting the expanded forms equal gives us:
$a^2 - b^2 = a^2 + 2ab + b^2$.

Step 3: Simplify and solve the equation:
- Subtract $a^2$ from both sides: $-b^2 = 2ab + b^2$.
- Add $b^2$ to both sides: $0 = 2ab + 2b^2$.
- Factor the right-hand side: $0 = 2b(a + b)$.

This gives us two possible conditions:
1) $2b = 0$, which implies $b = 0$.
2) $a + b = 0$, which implies $a = -b$.

Since $a = -b$ satisfies the equation for any $a$ if $b$ is not zero, and when $b = 0$, the equation simplifies to $0 = 0$, both conditions are valid.

Therefore, the solutions are $a = -b$ or $b = 0$.

In conclusion, the answer is: $a=-b$ or $0=b$.

To solve this problem, we will follow these steps:

• Step 1: Expand $(a+3b)^2$.
• Step 2: Expand $(3b-a)^2$.
• Step 3: Subtract the result of step 2 from step 1.

Now, let's work through the calculations:

Step 1: Expand $(a+3b)^2$
Using the formula $(x + y)^2 = x^2 + 2xy + y^2$, we let $x = a$ and $y = 3b$ to get:
$(a+3b)^2 = a^2 + 2 \cdot a \cdot 3b + (3b)^2$
$= a^2 + 6ab + 9b^2$.

Step 2: Expand $(3b-a)^2$
Again using the squaring formula, letting $x = 3b$ and $y = -a$, we have:
$(3b-a)^2 = (3b)^2 - 2 \cdot 3b \cdot a + a^2$
$= 9b^2 - 6ab + a^2$.

Step 3: Perform the subtraction
We subtract the expansion of $(3b-a)^2$ from $(a+3b)^2$:
$(a^2 + 6ab + 9b^2) - (9b^2 - 6ab + a^2)$
$= a^2 + 6ab + 9b^2 - 9b^2 + 6ab - a^2$
= $12ab$.

The solution to the problem is $12ab$, which corresponds to choice 2.

Let's examine the given equation:

$(x+3)^2=(x-3)^2$First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

$(a\pm b)^2=a^2\pm2ab+b^2$,

We'll start by opening the parentheses on both sides simultaneously using the perfect square formula mentioned, then we'll move terms and combine like terms, and in the final step we'll solve the resulting simplified equation:

$(x+3)^2=(x-3)^2 \\ \downarrow\\ x^2+2\cdot x\cdot3+3^2= x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9= x^2-6x+9 \\ x^2+6x+9- x^2+6x-9 =0\\ 12x=0\hspace{6pt}\text{/}:12\\ \boxed{x=0}$Therefore, the correct answer is answer A.

We know that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's look at triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

We substitute in our known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

Finally, we will add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$

To solve this problem, we'll follow these steps:

• Identify the given information.
• Use the geometric properties of a circle and a right triangle to find the radius.
• Express the area of the circle in terms of the given expression.

Now, let's work through each step:

Step 1: Given a circle with center $O$ and radii $AO$ and $OB$ such that $AO\perp OB$, each is a radius $r$, and $AB = \text{and}+2$.

Step 2: By the Pythagorean theorem, we know:

Step 3: Solving for the area of the circle:

The radius $r$ can be expressed by rearranging:

The area of the circle using this radius is:

Therefore, the expression for the area of the circle is $\frac{\pi}{2}[y^2+4y+4]$.

To solve this problem, let's expand and simplify each side of the given equation:

• Step 1: Expand the left side:
  • Expand $(x + 3)^2 = x^2 + 6x + 9$
  • Expand $(2x - 3)^2 = 4x^2 - 12x + 9$
  Combine them to get $x^2 + 6x + 9 + 4x^2 - 12x + 9 = 5x^2 - 6x + 18$.
• Step 2: Expand the right side:
  • Expand $5x(x - \frac{3}{5}) = 5x^2 - 3x$
• Step 3: Set the equations from both sides equal and simplify: $5x^2 - 6x + 18 = 5x^2 - 3x$
• Step 4: Subtract $5x^2$ from both sides: $-6x + 18 = -3x$
• Step 5: Simplify to: $-6x + 3x = -18$ or equivalently $-3x = -18$
• Step 6: Solve for $x$: $x = \frac{-18}{-3} = 6$

Therefore, the solution to the problem is $x = 6$.