(х+1)(х-1)+7

To simplify the expression given, which is \((x+1)(x-1) + 7\), we can start by using the FOIL (First, Outer, Inner, Last) method to expand the product of the two binomials \((x+1)(x-1)\). Here’s how it’s done:

\[ (x + 1)(x - 1) = x(x) + x(-1) + 1(x) + 1(-1) \]

This simplifies to:

\[ x^2 - x + x - 1 \]

Notice that the \(-x\) and \(+x\) cancel each other out. So we are left with:

\[ x^2 - 1 \]

Now, include the \(+7\) to get the full expression:

\[ x^2 - 1 + 7 \]

Combine the like terms, which are the constants -1 and +7:

\[ x^2 + 6 \]

So, the simplified form of the given expression \((x+1)(x-1) + 7\) is:

\[ x^2 + 6 \]