中四附加數(不等式)

x^2+(1-q)x+2q > 2
x^2+(1-q)x+2(q-1) > 0
判別式 = (1-q)^2 - 4(2(q-1)) < 0
q^2 - 2q + 1 - 8q + 8 < 0
q^2 -10q + 9 < 0
(q-1)(q-9) < 0
1<q<9
2. y=(5x^2+8x+4)/(x^2+x)﹐
yx^2 + yx = 5x^2+8x+4
(y-5)x^2 + (y-8)x - 4 = 0
由於x是實數,
判別式 = (y-8)^2 - 4(y-5)(-4) >= 0
y^2 - 16y + 64 + 16y - 80 >= 0
y^2 - 16 >=0
(y-4)(y+4) >= 0
y <= -4 或 y >= 4
3. 設y = (6x-1)/(3x^2-2x+1)
3yx^2 -2yx + y - 6x+1 =0
3yx^2 - (2y+6)x + (y+1) = 0
由於x是實數,
判別式 = (2y+6)^2 - 4(3y)(y+1) >= 0
4y^2 + 24y + 36 -12y^2 - 12y >= 0
-8y^2 +12y+36 >=0
2y^2 - 3y - 9 <=0
(2y+3)(y-3) <=0
-3/2 <=y <= 3
所以, -3/2 <= (6x-1)/(3x^2-2x+1) <= 3

x^2+(1-q)x+2q &gt; 2
x^2+(1-q)x+2(q-1) &gt; 0

判別式 = (1-q)^2 - 4(2(q-1)) &lt; 0
q^2 - 2q + 1 - 8q + 8 &lt; 0
q^2 -10q + 9 &lt; 0
(q-1)(q-9) &lt; 0
1&lt;9
2. y=(5x^2+8x+4)/(x^2+x)﹐
yx^2 + yx = 5x^2+8x+4
(y-5)x^2 + (y-8)x - 4 = 0

由於x是實數,
判別式 = (y-8)^2 - 4(y-5)(-4) &gt;= 0
y^2 - 16y + 64 + 16y - 80 &gt;= 0
y^2 - 16 &gt;=0
(y-4)(y+4) &gt;= 0
y &lt;= -4 或 y &gt;= 4
3. If y = (6x-1)/(3x^2-2x+1)
3yx^2 -2yx + y - 6x+1 =0
3yx^2 - (2y+6)x + (y+1) = 0
由於x是實數,

判別式 = (2y+6)^2 - 4(3y)(y+1) &gt;= 0
4y^2 + 24y + 36 -12y^2 - 12y &gt;= 0
-8y^2 +12y+36 &gt;=0
2y^2 - 3y - 9 &lt;=0
(2y+3)(y-3) &lt;=0
-3/2 &lt;=y &lt;= 3

so, -3/2 &lt;= (6x-1)/(3x^2-2x+1) &lt;= 3

中四附加數(不等式)

1)若對於x的一切實數值﹐二次式x^2+(1-q)x+2q恒大於2﹐試求q值的範圍。

1.) x^2+(1-q)&gt;2
x^2+(1-q)-(1-q)&gt;2-(1-q)
x^2&gt;2-1+q
x^2&gt;1+q
x^2-1&gt;1+q-1
q&gt;x^2-1 ,,

2)已知y=(5x^2+8x+4)/(x^2+x)﹐若x為實數﹐試求y值的範圍。

2.) y=(5x^2+8x+4)/(x^2+x)
y=x(5x+8)+4/x(x+1)
y=5x+8/x+1 ,,

3)若x為實數﹐試求數式(6x-1)/(3x^2-2x+1)的值的範圍

3.) (6x-1)/(3x^2-2x+1)
(6x-1)/x(3x-2)+1
(6x-1)/x(3x-2)+1
2+0.5/x+1
2.5/x+1 ,,

1)
x^2+(1-q)x+2q-2&gt;0
Since the equation is &gt;2, so it has no real roots because it does not touch the x-axis.
delta&lt;0
b^-4ac&lt;0
(1-q)^2-4*1*(2q-2)&lt;0
1-2q+q^2-8q+8&lt;0
q^2-10q+9&lt;0
(q-1)(q-9)&lt;0
use (x-a)(x-b)&lt;0 where(a