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1、分解因式單元測試一、填空題(每小題2分,共20分)1. .用提公因式法分解.3a2b+6ab2時,所提的公因式是2. .y2+x2分解因式的結(jié)果是 :3. 4m2.4mn+n2分解因式的結(jié)果是.9231 24. x x +=(x -).25. x2 + 9 =(x+3)2 .6. a4 (3b + c)2 = (a2 + 3b + c)().7. x5y2-xy2分解因式的結(jié)果是 :8. 若9x2 +kx +25是完全平方式,則k=.9. 若2 =,b=馬,那么(a + b)2(a - b)2 =.641510. 16a4 =(4+a2)()().二、選擇題(每小題3分,共30分)1 .把2(
2、x-3)+x(3-x)提取公因式(x-3)后,另一個因式是()A. x-2 B. x+2C. 2-x D. -2-x2 .下列各式分解因式結(jié)果正確的是()A5303222/32A. x y -2x y +x y = x y(x y -2xy)B . 6abc-8a2b2 =2abc(3-4abc)C . 4xy -8x2y2 =4xy(1-2xy)22D. xy+7xyy=y(x +7x)3.下列從左到右的變形哪個是分解因式()A. x2 -8x +16 = (x -4)2B . 3x +3y =6xyC. -a +b = -(b - a)D. 10x2 -5x-1 =5x(2x-1)-14
3、.將4x2-9分解因式的結(jié)果是()A. (4x+3)(4x-3) B. (2x+3)2C. (2x -3)2D. (2x+3)(2x-3)5 .下列各式中能用平方差公式分解的是()A 22222222.-x y B. m +(n)C. 16a -81b D. - (-x) (x+y)6 .下列各式中能用完全平方公式分解的是()22122A. a+9 B. x+x+ C. x4y D.x+2x+4 47 .x416, x2 +4-4x的相同因式是()A.x2+4 B. x2 -4 C. x+2D. x-28 .下列因式分解正確的是()A. 2x2 -3xy -x =x(2x -3y)B . (x
4、2 +1)2 4x2 =(x2 +2x +1)(x2 2x +1)C . -x2 +9y2 =(3y +x)(3y -x)D . 3(x -1)y - (1 -x)z =(x - 1)(3y -z)9 . 9(a+1)2-16b2分解因式的結(jié)果為()A. (3a+4b+3)(3a + 4b-3) B. (3a+4b+3)(3a + 4b+3)C. (3a+4b+3)(3a 4b-3) D. (3a+4b+3)(3a 4b+3)10 .若 a2 +ma +1 = (a -1)2 ,貝U m 的值為()9 3 2 2A. 2 B. 3 C. - D. 233三、解答題(本大題共50分)1 .(本題
5、8分)判斷并改錯:(1) 16x2 -4y2 =(4x)2 -(2y)2 = 2(2x + y)(2x - y);(2) x2 +2x+4 = (x+2)2.2 .(本題16分)分解因式:2 24(1) 9ab - b ;(2) 4(a +2b)2 -25(a -b)2 ;(3) ;x2-2xy+2y2;(4) ab2 +2ab +a .3 .(本題10分)計算:(1 ) 3202 -400 -340298; 982 +4 +98父4.4 .(本題6分)已知a +2b=3,求1a2+4 b2+4ab的值. 9995 .(本題10分)正方形甲的周長比正方形乙的周長長96cm,它們的面積相差960
6、 cm2,求這兩個正方形的邊長.四、探索題(每小題10,共20分)1 .已知多項式ax2+bx+1可分解成一個一次多項式平方的形式.(1)請寫出一組滿足條件的a、b的整數(shù)值;(2)猜想出a、b之間的關(guān)系,并表示出來.2 .計算:32-12=, 52一32=, 72 -52 =, 92 -72 = .(1)根據(jù)以上的計算,你發(fā)現(xiàn)了什么規(guī)律?請用含有n的式子表不;(2)用分解因式的知識說明你發(fā)現(xiàn)的規(guī)律.參考答案2.3.4.5.一、填空題(x y)(x- y)/22(m - n)I346x6. a2 -3b -c7. xy 2(x2 +1)(x +1)(x -1)土曰5 22 2.42 . 222
7、. 2提木:x y -xy = xy (x 1) = xy (x +1)(x 1) = xy (x + 1)(x + 1)(x -1).8. 30提示:9x2 +kx +25 = (3x5)2 =9x2 30x +25 , k =30 .9. 12提示: (a b)2 -(a -b)2 =(a b) (a - b)( a b)-(a - b) = 4ab=4且空64 15210. 2+a, 2-a二、選擇題1. C提示:2(x -3) x(3 -x) = 2(x -3) - x(x - 3) - (x - 3)(2 - x)2. C3. A4. D5. C6. B7. D提示:x4 -16 =
8、(x2 +4)(x2 -4) =(x2 +4)(x +2)(x -2),x2 +4-4x = x2 - 4x +4 = (x - 2)2 .故它們的相同因式是x -28. C9. D10. C提示: 由 a2 + ma +1 = (a -)2,得 a2 + ma +- = a2 -2 a +-,所以 m =. 939393三、解答題1 .解:(1)錯,應(yīng)為:16x24y2= 4(4x2y2)=4(2x +y)(2x y); (2)錯,此題無法分解.2 .解:(1) 9a2b2 b4 = b2(9a2 b2) =b2(3a+b)(3a b);(2) 4(a +2b)2 -25(a -b)2_22
9、= 2(a 2b) -5(a-b)= 2(a 2b) 5(a - b) 2(a 2b) -5(a -b)= 2a 4b 5a -5b 2a 4b - 5a 5b=-3(7a -b)(a -3b);/C122 _ 12212(3) -x -2xy +2y =-(x -4xy+4y ) = -(x-2y); 222(4) ab2 +2ab+a = a(b2 +2b +1) = a(b+1)2.3 .解:(1) 3202 -400 -340 x2982_2_= 320 -20 -340 298=(320 20)(320 - 20) - 340 298= 340 300 -340 298= 340x(
10、300-298) =680 ;(2) 982 +4 +98 父4 = 982 + 2 父98M 2 +22 =(98 +2)2 =1002 =10000 .1 24 24122124 . 解:-a +b +-ab = - (a +4ab+4b )=(a+2b).9991919由 a +2b = 3,得 1(a+2b)2 =_ 父32 =1 . 995 .解:設(shè)正方形甲的邊長為xcm,正方形乙的邊長為 ycm (xy),則;4x-4y=96,22、x2 -y2 =960.由,得x-y = 24.由,得 x2 y2 = (x + y)(x y) = 960 ,即 24(x + y)=960,所以x+y=40. 由解得x=32, y = 8.答:正方形甲的邊長為32cm,正方形乙的邊長為8cm.四、探索題1 . (1)如 a = 9, b = 6 b2 =4a2 .(1)根據(jù) 32 12 = 8 , 52 32 = 8父 2 , 72 52 =8x3 , 92 72 = 8父4 , 發(fā)現(xiàn)(2n +1)2 (2n 1)2 =8n (n 為正整數(shù));(2)說明:左邊=(2n+1) +(2n 1)(2n+1) (2n 1) = 4n 父2 =8n =右 邊.