It is known that the expansion of (x-m) (X & # 178; + NX + 1) does not contain x, X2 terms. Try to find the value of M, n

It is known that the expansion of (x-m) (X & # 178; + NX + 1) does not contain x, X2 terms. Try to find the value of M, n

=x³+nx²+x-mx²-mnx-m
=x³+(n-m)x²+(1-mn)x-m
∵ does not contain x, X & # terms
∴n-m=0
1-mn=0
∴m=1 m=-1
n=1 n=-1

If the length m, N, P of three line segments satisfies P & # 178; = M & # 178; - N & # 178;, the triangle formed by these three line segments is a right triangle?

It's a right triangle
p²=m²-n²
↓ (move - N & # 178; to the left of the equal sign)
p²+n²=m²
The Pythagorean theorem is obtained, that is, in a right triangle, the sum of the squares of the two right angles equals the square of the hypotenuse

Let AB be the length of two right angle sides of a right triangle (A & # 178; + B & # 178;) (A & # 178; + B & # 178; + 1) = 12
Let AB be the length of two right angles of a right triangle (A & # 178; + B & # 178;) (A & # 178; + B & # 178; + 4) = 21
Let AB be the length of two right angle sides of a right triangle (A & # 178; + B & # 178;) (A & # 178; + B & # 178; + 4) = 21, and find the oblique side length
How to solve the problem of X (x + 4) = 21?

So let a & # 178; + B & # 178; = x, then x (x + 4) = 21, then (x-3) (x + 7) = 0, then x = - 7 or x = 3, because a & # 178; + B & # 178; = x, so x is greater than or equal to 0, so x = 3, then C & # 178; = 3, so C = (√ 3)