Question

解下列方程。请各位高手帮一下忙。急需！！ 过程也要写。

①：（x+5）^2=3（x+5） ②：（x-4）^2-x+4=0 ③：3(x+1)(x-2)=2(X^2-1)

Answer 1

1、x+5=0或不为0
是0时x=-5;不为0，两边消去x+5，得x=-2
2，同理1，x=4或者x=5
3，同理x^2-1=(x-1)(x +1)
当x+1=0时或不为0
得x=-1或x=4

追问

4(x-3)^2+x(3-x)=0 (3x-1)^2=4(2x-3)^2
这两题怎么做啊？

追答

4(x-3)^2+x(3-x)=0这个跟前几个一样啊，移项
4(x-3)^2=x（x-3）得到x=3或x=4
这个 (3x-1)^2=4(2x-3)^2
就需要展开了9x^2-6x+1=16x^2-48x+36
7x^2-42x+35=0
x^2-6x+5=0
x=1 x=5

Answer 2

搞定了没?
没搞定的话找我吧
happyq6
happyq6@163.com

放6个Label,1个Command,1个Text控件数组（0到3）
然后拷贝以下代码
Const T = 813
Const C = 20
Const Gama = 65000000
'LMP = T(C + log¦Ó)
'LMP = a0 + a1* (log¦Ò) + a2* (log¦Ò) 2 + a3* (log¦Ò) 3
'ÆäÖÐ C=20 £¬ a0 ¡¢a1 ¡¢a2 ¡¢a3 Ϊ¸ø¶¨³£Êý
'¦Ò= 65MPa , T = 540 ¡æ
'Çóδ֪Êý¦Ó
Private Sub Command1_Click()
Dim a(3) As Single, LMP As Single, temp As Single, Tuo As Single, str As String
LMP = 0
str = "Follow the formula:" & vbCrLf
str = str & "LMP = T(C + log¦Ó)" & vbCrLf
str = str & "LMP = a0 + a1* (log¦Ò) + a2* (log¦Ò) 2 + a3* (log¦Ò) 3" & vbCrLf & vbCrLf
temp = Log(Gama)
str = str & "Log¦Ò=" & temp & vbCrLf
For i = 0 To 3
a(i) = Val(Text1(i).Text)
LMP = LMP + a(i) * temp ^ i
Next
str = str & "a0=" & a(0) & " a1=" & a(1) & " a2=" & a(2) & " a3=" & a(3) & vbCrLf
str = str & "LMP=" & LMP & vbCrLf
temp = LMP / T
str = str & "LMP/T=" & temp & vbCrLf
temp = temp - C
str = str & "LMP/T-C=" & temp & vbCrLf
Tuo = Exp(temp)
Tuo = Format(Tuo, ".00E")
str = str & "¦Ó=Exp(LMP/T-C)=" & Tuo
Label6.Caption = str
End Sub
Private Sub Form_Load()
Text1(0).Text = ""
Text1(1).Text = ""
Text1(2).Text = ""
Text1(3).Text = ""
Label1.Caption = "a0="
Label2.Caption = "a1="
Label3.Caption = "a2="
Label4.Caption = "a3="
Label6.Caption = ""
Command1.Caption = "Calculate"
Label5.Caption = "T=540 Degree C =813K" & vbCrLf & vbCrLf & "¦Ò=65 MPa" & vbCrLf & vbCrLf & "C=20"
End Sub
我觉得T应该计算为绝对温度
Gama应该为Pa单位
如果你觉得应该用别的单位改改数据就可以了
要程序的话就mail me吧
法给你

Answer 3

1.两边都有x+5看见没？说明有一解为-5 然后约掉 x+5=3 得 另一解-2
你写的时候不要直接约 先移项 提取公因式 注意规范啊

第二题和第一题一样的题型 解为4和5 弄不懂再问我
第三题等式两边展开 因式分解得（x+1）（X-4）=0 解为 -1和4