maths..2次方程
用取平方根方法解下列方程
1.2(3+x)^2=32
2.1/3(4-x)^2=27
3.(x-4)^2-17=0
4.3(3x+1)^2+5=0
唔識做..教教我thx"要有式><
回答 (3)
1. 2(3+x)^2 = 32
(3+x)^2 = 32 / 2
(3+x)^2 = 16
(3+x) = 開方16
3+x = +4 or 3+x = -4
x= +1 or x = -7
2.1/3(4-x)^2=27
(4-x)^2=27x3
(4-x)^2=81
(4-x)=開方81
(4-x)= +9 or -9
4-x = +9 or 4-x =-9
x= 4-9 or 4+9 =x
x = -5 or x = 13
3.(x-4)^2-17=0
(x-4)^2 = 17
(x-4) = 開方17
x = +_開方17 + 4
4.3(3x+1)^2+5=0
3(3x+1)^2 = -5
(3x+1)^2 = -5/3
no solution
因為開方入面唔會是負數的
1.
2(3+x)^2=32
(3+x)^2=16
3+x=4
x=1
2.
1/3(4-x)^2=27
(4-x)^2=81
(4-x)=9
x=-5
3.
(x-4)^2-17=0
x^2-8x+16-17=0
x^2-8x-1=0
x=(8+開方68)/2 或 (8-開方68)/2
4.
3(3x+1)^2+5=0
3(9x^2+6x+1)+5=0
27x^2+18x+8=0
no real solution
參考: me
收錄日期: 2021-04-23 23:05:39
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