初二數學題高分懸賞急

時間 2022-10-21 09:31:53

1樓:

見圖公式編輯器做的花了點時間

2樓:匿名使用者

(a²+3)/(a²-1)-(a+1)(a-1)+1=[a²+3-(a+1)²+a²-1]/(a²-1)=(a²-2a+1)/(a²-1)

=(a-1)²/(a+!)(a-1)

=(a-1)/(a+1)

原式=1/(x-2)²+1/[2(x+2)]-x/[2(x+2)(x-2)]

=1/(x-2)²+(x-2-2x)/[2(x+2)(x-2)]=1/(x-2)²-(x+2)/[2(x+2)(x-2)]=1/(x-2)²-1/[2(x-2)]

=[2-(x-2)]/[2(x-2)²]

=(4-x)/[2(x-2)²]

原式=×(xy)/(y-1)

=[1/(x-y)-y/(x-y)]×(xy)/(y-1)=(1-y)/(x-y)×(xy)/(y-1)=xy/(y-x)

3樓:老黃知識共享

第一題原式=[(a^2+3)-(a^2+2a+1)]/(a^2-1)-1=(-2a+2)/(a^2-1)-1=-2/(a+1)-1=(-2-a-1)/(a+1)=-(a+3)/(a+1)

第二題原式=1/(x-2)^2-x/[(x+2)(x-2)]+1/[2(x+2)]

=2(x+2)/[2(x+2)(x-2)^2]-2x(x-2)/[2(x+2)(x-2)^2]+(x-2)^2/[2(x+2)(x-2)^2]

=[2(x+2)-2x(x-2)+(x-2)^2]/[2(x+2)(x-2)^2]

=(2x+4-2x^2+4x+x^2-4x+4)/[2(x+2)(x-2)^2]

=(-x^2+2x+8)/[2(x+2)(x-2)^2]

=-(x+2)(x-4)/[2(x+2)(x-2)^2]

=-(x-4)/[2(x-2)^2]

第三題原式=*xy/(y-1)

=[1/(x-y)-y/(x-y)]*xy/(y-1)

=-(y-1)/(x-y)*xy/(y-1)

=-xy/(x-y)

4樓:匿名使用者

(a²+3)/(a²-1)-(a+1)/(a-1)+1

=(a²+3)/(a+1)(a-1)-(a+1)²/(a+1)(a-1)+(a²-1)/(a+1)(a-1)

=(a²+3-a²-2a-1+a²-1)/(a+1)(a-1)

=(a²-2a+1)/(a+1)(a-1)

=(a-1)²/(a+1)(a-1)

=(a-1)/(a+1)

1/(x²-4x+4)-x/(x²-4)+1/(2x+4)

=1/(x-2)²-x/(x-2)(x+2)+1/2(x+2)

=[2(x+2)-2x(x-2)+(x-2)²]/2(x+2)(x-2)²

=(2x+4-2x²+4x+x²-4x+4)/2(x+2)(x-2)²

=(-x²+2x+8)/2(x+2)(x-2)²

=-(x-4)(x+2)/2(x+2)(x-2)²

=-(x-4)/2(x-2)²

[(x-y)/(x²-2xy+y²)-(xy+y²)/(x²-y²)]*xy/(y-1)

=[(x-y)/(x-y)²-y(x+y)/(x-y)(x+y)]*xy/(y-1)

=[(1/(x-y)-y/(x-y)]*xy/(y-1)

=(1-y)/(x-y)*xy/(y-1)

=-xy/(x-y)

5樓:好

第一題(a^2+3)/(a^2-1)-(a+1)/(a-1)+1

=(a^2+3)/(a^2-1)-(a+1)(a+1)/(a-1)(a+1)+(a^2-1)/(a^2-1)

=(a^2+3)/ (a^2-1)-(a+1)^2/(a^2-1)+(a^2-1)/(a^2-1)

=(a^2+3-a^2-2a-1+a^2-1)/(a^2-1)

=(a^2-2a+1)/(a^2-1)

=(a-1)^2/(a+1)(a-1)

=(a-1)/(a+1)

第二題1/(x^2-4x+4)-x/(x^2-4)+1/(2x+4)

=1/(x-2)^2+x/(x-2)(x+2)-1/2(x+2)

=[2(x+2)+2x(x-2)-(x-2)^2]/2(x+2)(x-2)^2

=x(x+2)/2(x+2)(x-2)^2

=x/2(x-2)^2

第三題[(x-y)/(x^-2xy+y^2)-(xy+y^2)/(x^2-y^2)]*xy/(y-1)

=[(x-y)/(x-y)^2-y(x+y)/(x+y)(x-y)]*xy/(y-1)

=[1/(x-y)-y/(x-y)]*xy/(y-1)

=[(1-y)/(x-y)][xy/(y-1)]

=-xy/(x-y).

6樓:

第一題=(a^2+3-a^2-2a-1+a^2-1)/a^2-1=(a-1)^2/(a-1)(a+1)

=a-1/a+1

第二題:

=[2(x+2)-2x(x-2)+(x-2)^2]/2(x-2)^2(x+2)

=(4-x)/2(x-2)^2

第三題:

=[(1/x-y)-(y/x-y)]*(xy/y-1)=xy/y-x

急,初二數學題

證明 如圖,連線ae ef垂直平分ab ae be acb的補角為110 acb 70 cad 20 adc 180 acb cad 90 又d為ce中點 ad垂直平分ce ac ae be ac 連線ae 角acb的補角是110 acb 70 cad 20 adc 90 ade 90 d為線段ce...

數學題,要過程,高分懸賞,急

1 an 2 n 1 tn 6sn 8n 2 bn tn t n 1 6sn 8n 6s n 1 8 n 1 2 6an 8 2 4 6 n 1 6n 2 2 cn bn 8n 3 cn 2n 1,d1 c1 3,d2 c d1 c3 2 3 1 7 2 3 1,d n 1 c dn 2 n 2 1...

初二數學題知道答案但很糾結絕對高分懸賞

b句本身沒錯,但不對題,c這句話不對,應該是兩個全等的等腰直角三角形構成正方形d句不對,軸對稱圖形不一定是正方形 故選a。該選a.c其實說法是個錯的。是你把答案弄錯了吧 我說a是對的吧。本來c就是錯誤的命題。兩個直角三角形一定能構成正方形這命題是錯誤的。a本身是對的,但並沒用到這個原理,你折出來對稱...

高分懸賞 數學題求解,高分懸賞!!求解一道數學題!!

1.簡便計算 15 888 32 0.125 0.25 10 888 5 888 8 4 0.125 0.25 8880 4440 1 13320 1 13321 1 2 3 4 5 6 2003 2004 2005 1 3 2 5 4 2005 2004 1 5 9 4009 1 4009 100...

急。初二的幾道數學題

第一題 左右兩式分母提取 x y 然後左邊分子分母同乘以y,右邊分子分母同乘以x。然後左右分子做加或差即可 第二題 a 2且a不等於 4,因為當a 4時,左邊恒為2。當然不成立啦 第三題 設跑到長度s,甲乙速度分別是v1,v2.易知s 15 v1 v2 s 40 v1 兩式相減,得到是s 24 v2...