The area of a rectangle is X & # 178; - Y & # 178;, and the area of a square with its long side as its side length is The area of a rectangle is X & # 178; - Y & # 178;, and the area of a square with its long side as its side length is I know the solution is as follows: X & # 178; - Y & # 178; = (x + y) (X-Y). When x > 0, Y > 0, the long side is x + y, so the area is (x + y) &# 178; = x & # 178; + 2XY + Y & # 178; But when x ＞ 0, y ＜ 0, the long side is (X-Y), so the area is (X-Y) &# 178; = x & # 178; - 2XY + Y & # 178; But this is a single choice question, and the answer is X & # 178; + 2XY + Y & # 178; what's wrong with the second idea?

The area of a rectangle is X & # 178; - Y & # 178;, and the area of a square with its long side as its side length is The area of a rectangle is X & # 178; - Y & # 178;, and the area of a square with its long side as its side length is I know the solution is as follows: X & # 178; - Y & # 178; = (x + y) (X-Y). When x > 0, Y > 0, the long side is x + y, so the area is (x + y) &# 178; = x & # 178; + 2XY + Y & # 178; But when x ＞ 0, y ＜ 0, the long side is (X-Y), so the area is (X-Y) &# 178; = x & # 178; - 2XY + Y & # 178; But this is a single choice question, and the answer is X & # 178; + 2XY + Y & # 178; what's wrong with the second idea?

What's wrong with the second idea? Because area is a scalar, not a vector. It has no negative value, and Y & # is always positive

The height of a rectangle is reduced by 2cm, the surface area is reduced by 48cm2, and the volume of a square is calculated

48÷2÷4=6
6×6×6=216

From the square iron sheet, take a rectangular strip 2cm wide, and the remaining area is 80cm2. What is the side length of the original square iron sheet?

Let the original side length be a
a×(a－2)＝80
The solution is a = 10

Cut a rectangular strip 2cm wide from the square iron sheet, and the remaining area is 80 square centimeters

No square has x sides
X²-2X=80
X-2X-80=0
(X-10)(X+8)=0
X1=10
X2 = - 8, rounding off
Square side length = 10