超難...數學高手先好入

2011-12-17 8:58 pm
need PROCESS!!!!!!!!!!!!!!!

1.Simplify 1/3(2y3 -4y-3)-1/3(y3+y+3y2)-(1/3y3-2/3y)

2.factorize xy+y+x=1

3.simplify (4x-3)(6-x)-x(x-2)

4.simplify 5/3(1/x)(x/3)

5.if the ordered pairs (m,0) and (0,n) satisfy the equation 4x-5y-20=0,find the sum of the values of m and n

6.if the circumference of a circle of radius R is equal to double of the perimeter of a square of side x,find the ratio of the area of the square to that of the circle
(Given that the area of the circle with radius R is 圓周率x R2

7.solve the equality 2x大過等於23x-6/7

回答 (2)

2011-12-18 3:16 am
✔ 最佳答案
1.
(1/3)(2y^3-4y-3) - (1/3)(y^3+y+3y^2) - [(1/3)y^3 - (2/3)y]
= (1/3)(2y^3 - 4y - 3 - y^3 - y - 3y^2 - y^3 + 2y)
= (1/3)(-3y - 3 - 3y^2)
= -(y^2 + y + 1)

2. (it seems can't be factorized)

3.
(4x-3)(6-x) - x(x-2)
= -4x^2 + 27x - 18 - x^2 + 2x
= -5x^2 + 29x - 18
= -(5x^2 - 29x + 18)

4.
(5/3)(1/x)(x/3)
= 5/9

{5/[3(1/x)]} (x/3)
= {5/[3/x]} (x/3)
= {5x/3} (x/3)
= 5x^2 / 9

5.
4(m) - 5(0) - 20 = 0 => m = 5
4(0) - 5(n) - 20 = 0 => n = -4
m + n = 1

6.
perimeter of square = 4x
circumference: 2(pi)(r) = 2 (4x) => x = (pi)r/4
area of ratio: (pi)r^2 / x^2 = (pi)(r^2) / [(pi^2) (r^2) / 16]
= 16/(pi)

7.
2x >= 23x - 6/7
6/7 >= 21x
2/49 >= x
2011-12-18 2:55 am
1.
1/3(2y3 -4y-3)-1/3(y3+y+3y2)-(1/3y3-2/3y)
=1/3(2y3 -4y-3)-1/3(y3+3y2+y)-1/3(y3-2y)
=1/3[(2y^3-4y-3)-(y^3+3y^2+y)-(y^3-2y)]
=1/3(2y^3-4y-3y-y^3-3y^2-y-y^3+2y)
=1/3(-3y^2-3y-3)
=-y^2-y-1
=-(y^2+y+1)

2.
xy+y+x=1
x+xy+y=1

3.
(4x-3)(6-x)-x(x-2)
=24x-4x^2-18x-3x-x^2+2x
=-5x^2+8x

4.
5/3(1/x)(x/3)
=5/9

5.
4x-5y-20=0
when x=m y=0,
4m-5(0)-20=0
m=20/4
m=5
when x=0 y=n,
4(0)-5n-20=0
-5n=20
n=-4
so
(m,n)=(4,-4)

6.
2丌R=2丌x^2
R=x^2

7.
2x≥23x-6/7
6/7≥23x-2x
6/7≥21x
(6/7)/21≥x
2/49≥x


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