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Help me solve some math problems!
1, (1)S=(50-X)X(CM) may need to be changed. Remember to follow the unit.

(2) When S=400 square centimeters, (50-X)X=400, so X 1=40X2= 10, 10 can be omitted, but x is set to be longer than the width, but the final answer is the same.

(3) When S=625 cm2, (50-X)X=625, so X 1=X2=25.

(4) If the maximum value of S cannot be 700 square centimeters, the matching method can be used to explain that the maximum value of S can only be 625, and the square of a number is non-negative.

In other words, when X=25, the rectangle is a square, and 25*25=625 is the maximum.

2. Suppose there are X students in the original Grade Three (1) class.

(x-1+6) * (x+6) =1260 The result is X 1=30X2=-4 1 (truncation).

3. (55-X)(45-X)=2000, and the solution is X 1=5X2=95 (truncation).

4. Because the height of the box is equal to cutting off the sides of four squares.

So let the height of the box be X(50-2X)(40-2X)=600.

So the solution of (4X- 140)(X- 10) is X 1= 10X2=35 (truncation).

5, because shaking hands is not with yourself but with two people.

Therefore, if there is a classmate X (X- 1)*X/2=300, the solution is X 1=25X2=-24 (abbreviated).

Finally, in fact, binary linear equations are not difficult, mainly in practical applications. The most important thing in practical application is that the value range of x should proceed from reality.