第1课时分式的乘除3.docx
“分式的乘除法”复习题一.填空题1 .约分:（n为正整数）m2+ 2m+1 _ 1 -m22 .计算：222/-) x4 / ab、2/。、2 /人、/ 1、4(2c)(-)=；(一-)()()=2cb a ab（一2 x -Ixyy2 x -y(xy-x ) +产二xyx二.判断题下列运算正确的打“ J ”，错误的打“ x ”:1.2x y x+ y 2x 2x;=r 1 =%+> %+y y x+yx + y2x +y)3z%6+ y3z33.z24-(. )4.2b"）（n为正整数）（）a2n5.2b33a238b9):可三.选择题1.已知x:y = 3,则分式x1的值是B.27261C.一2D.14132：在分式q,3x6孙2x + 2yy 一%(x y)23 3x + y2 2 % -y中，最简分式有（）A. 1个B. 2个C.3个D.+ 2x +1B. 72x x 1%-13 .下列各式正确的是（22x _yA r x + y x+y八 a a + bc.=b 2bD.-)2 * 4 a + ba2+b26.+ X +11 -x2+l " 1 + x2+ (x2 + x 2)四.计算1 . 8/(兴)(一半).4y x y2 .(三)3 (/)&£)2 a + bb + aCOy +(a + b)y + ab a -3 Z , 2 n-x -(a + b)x + ab b -y 6x + 9 9x + 34；x 6 3x 10 2(x - 5)oao+7。+ 104 +1-4 + 1.!a2 + 6a+ 5 a2 + 4。+ 4 a + 297 9小“一。一、2/。”8.(>.(yab a-b(力，)3(q2+2qx + /)2i2744z 22 2a +xa x (q ax +)a2 +c2 2ab -b2c2 + 2bca1 b2 + c 2ac ci + Z? c + 2abLI.(x+y)2孙+ )2孙一 V (X y)212.1 y-x321 1-x + y3213.+ 2/ + x + 2 4x + 4+1：+ 4x + 4， 2x - x + 2 龙-114.2x 16 12-4% , 、 + (x + 2)厂4x + 4 x -415.01b23(a + b)(a2-b2)a2cr (b - a)I3,其中q=L, b=-o 24c 2 八b + 4ab2 +16 (8/-2) + -2b2 -5ab-3a2【试题答案】1.m + 1i；a + i；,2. 4a2b4c21ab427/b1.三.1.四.2.4.5.2.3. B1.解：原式=8一3x476z _ 36xz y42.27 r9解：原式=-(a-b)(a2 +a(a + b)327( - b)(a2 +ah + b2 )x9(4 + Z7)(/? I2)?3.解:原式=（y + a）（y + b）（a + x）（。-）(x-q)(x-h) (b+ y)(b- y)(y + a)(x + a) _ (y + ax + a)(x_/?)(/?_ y) (x-by- b)4.解:原式=(x (x 3)(x + 2)(x-5)(x + 2)(x + 3)(x-3)x + 3 i2(x-5) - 25.解:原式=2(x-3) 1(x 2)2 x + 3(x + 3)(x 2)3-x6一解:原式=+ X + 1(1 X)(%2 + X + 1)7.解:8.解:9.解:X2 +11 + x2(x + 2)(x-l)原式=(a + 2)(a + 5)(a+ 1)(+ 5)(4 + 2)2原式=原式=(a + b)2 (a b)?a3 /( +2 a(a + b)2a2b2(a + x)3(a-x)3226r +x(a-b)3a2b2(6Z2+ x2)(a + x)(a- x)(6Z + X)4b2(a-x)in到西卜(a-b + c)(a-b-c)10,. 解：原式=( + /7_ c)(q b c)(a + b + c)(a + b-c) _ a + b + c(a + b- c)(a -b + c) a + b-c(x+y)(xy)ii.解：原式: s+y)2 y (x - y)y (% + y)12.2x-3y62x + 3y62x + 3y解：原式二个.勺2x-3yx_ 2x + 3y 2x-3y _2x-3y 2x + 3y13.解：原式=(x + 2) + (x + 2)(x 2)2(x + l)(x 1)14.15.16.(x + 2)2解：原式二2(x 8J(x 2)2%2(x 2) (x 2)%2 +1(x + 2)(x-2)14(3 - x) x + 2 2(3 x)(x 2)01b2 解：原式=巴， 3( + b)( + 加4( 一力4a18a6(b d)3S(a + b)3b2(a-b)3a2解：原式=2(2 + b)(2-b)(2b + a)(b 2a)b(2a + b)2 (b 2a)(b 3a)当 =b ='时，原一式=一2x(一g) + 4x J2x(-) + 24代数式的值为022x 6x x 65. + (x + 3)4 4x + x3 x
