(1)1-1/(n+1)
(2)1/n-1/(n+2)=2/[n(n+2)]
(3)1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)=3/[2x+18]
1/x-1/(x+9)=3/[2x+18]
9/[x(x+9)]=3/[2x+18]
9*[2x+18]=3*[x(x+9)]
18x+162=3x^2+27x
3x^2+9x-162=0
(3x-18)(x+9)=0
x=6或x=-9
經(jīng)檢驗,x=-9是原方程的增根,舍去.
所以,原方程的根為x=6.
探索規(guī)律題
探索規(guī)律題
觀察下列等式:1/(1*2)=1-1/2,1/(2*3)=1/2-1/3,1/(3*4)=1/3-1/4
以上三個等式兩邊分別相加得:
1/(1*2)+1/(2*3)+1/(3*4)=1-1/2+1/2-1/3+1/3-1/4=1-1/4=3/4
(1)直接寫出下列各式的計算結(jié)果:
1/(1*2)+1/(2*3)+1/(3*4)+……+1/[n(n+1)]=_______
(2)猜想并寫出:1/[n(n+2)]=_______
(3)探究并解方程:
1/[x(x+3)]+1/[(x+3)(x+6)]+1/[(x+6)(x+9)]=3/[2x+18]
觀察下列等式:1/(1*2)=1-1/2,1/(2*3)=1/2-1/3,1/(3*4)=1/3-1/4
以上三個等式兩邊分別相加得:
1/(1*2)+1/(2*3)+1/(3*4)=1-1/2+1/2-1/3+1/3-1/4=1-1/4=3/4
(1)直接寫出下列各式的計算結(jié)果:
1/(1*2)+1/(2*3)+1/(3*4)+……+1/[n(n+1)]=_______
(2)猜想并寫出:1/[n(n+2)]=_______
(3)探究并解方程:
1/[x(x+3)]+1/[(x+3)(x+6)]+1/[(x+6)(x+9)]=3/[2x+18]
數(shù)學人氣:532 ℃時間:2020-06-29 07:36:19
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