幾條簡單f.4數學題...

2007-04-26 3:46 am
1.) f(x) = x^2 - 3 / 2x + 7
 a.) f(1/4)
 b.) f(2/5)

2.) 令 r 成為 t = y(r - 4) + 2r 的主項

3.) -3(2x + 7) - 8 > 4 - x

4.) 8(2 - 5x) > 2x + 3


help..

回答 (5)

2007-04-26 3:57 am
✔ 最佳答案

1.) f(x) = x^2 - 3 / 2x + 7
 a.) f(1/4)
f(x) = x^2 - 3 / 2x + 7
f(0.25)=0.0625-3/0.5 +7
=0.0625-6+7
=1.0625

 b.) f(2/5)
f(x) = x^2 - 3 / 2x + 7
f(0.4) =0.16 - 3 / 0.8 + 7
=-3.59+7
=3.41


2.) 令 r 成為 t = y(r - 4) + 2r 的主項
t = y(r - 4) + 2r
t=yr-4y+2r
t+4y=r(y+2)
r=(t+4y)/(y+2)


3.) -3(2x + 7) - 8 > 4 - x
-3(2x + 7) - 8 > 4 - x
-6x-21-8>4-x
-5x>33
x<-6.6


4.) 8(2 - 5x) > 2x + 3
8(2 - 5x) > 2x + 3
16-40x>2x+3
13>42x
x<42/13
2007-04-26 4:04 am
1.a)
f(x)=x^2-3/2x+7
f(1/4)=(1/4)^2-3/2(1/4)+7
=1/16-3/8+7
=107/16

b.)
f(2/5)= (2/5)^2-3/2(2/5)+7
= 4/25-3/5+7
= 164/25

2).
t=y(r-4)+2r
t=ry-4r+2r
t=ry-2r
t=r(y-2)
t/(y-2)=r
r = t/(y-2)

3).
-3(2x+7)-8 &gt; 4-x
-6x-21-8 &gt; 4-x
-6x+x &gt; 4+29
-5x &gt; 33
x &lt; -33/5

4.)
8(2-5x) &gt; 2x+3
16-40x &gt; 2x+3
-40x-2x &gt; 3-16
-42x &gt; -13
x &lt; 13/42
參考: me
2007-04-26 4:02 am
1.) a.) f(1/4)=1/16-3/8+7
=107/16

b.)f(2/5)=4/25-3/5+7
=164/25

2.) t=yr-4y+2r
t+4y=r(y+2)
r=(t+4y)/(y+2)

3.) -6x-21-8&gt;4-x
-33&gt;5x
x&lt;-33/5

4.) 16-40x&gt;2x+3
13&gt;42x
x&gt;13/42
參考: 自己
2007-04-26 3:57 am
1.) f(x) = x^2 - 3 / 2x + 7
a.) f(1/4) = 0.25^2 - 3/(2x0.25) + 7 = 1.0625
b.) f(2/5) = 0.4^2 - 3/(2x0.4) + 7 = 3.41

2.)
t=y(r-4)+2r
t=yr-4y+2r
t=r(y+2)-4y
t+4y=r(y+2)
r= (t+4y)/(y+2)

3.) -3(2x + 7) - 8 &gt; 4 - x
-6x - 21 -8 &gt; 4 - x
-33&gt;5x
x&lt;-33/5

4.) 8(2 - 5x) &gt; 2x + 3
16-40x&gt;2x+3
13&gt;42x
x&lt;13/42
2007-04-26 3:57 am
1. a. f(1/4) =(1/4)^2-(3/2)(1/4)+7
=1/16-3/8+7
=107/16
b. f(2/5) = (2/5)^2-(3/2)(2/5)+7
=4/25-6/10+7
=164/25

2, t = y(r - 4) + 2r
t- y(r - 4) = 2r
t- y(r - 4)
----------- = r
2
參考: me


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