alimov-9-gdz (Алгебра - 9 класс - Алимов)

Домашняя работапо алгебреза 9 класск учебнику «Алгебра. Учебник для 9 кл.общеобразовательных учреждений» Ш.А. Алимов,Ю.М. Колягин, Ю.В. Сидоров и др. — 6-е изд. — М.:«Просвещение», 2001 г.учебно-практическоепособие3СОДЕРЖАНИЕСтепень с рациональным показателем ........... 4ГЛАВА IV. Элементы тригонометрии ..........

73ГЛАВА V. Прогрессия .................................... 1194СТЕПЕНЬС РАЦИОНАЛЬНЫМ ПОКАЗАТЕЛЕМ62.1) 23 + ( – 3)3 – ( – 2)2 + ( – 1)5 = 8 + ( – 27) – (4) + ( – 1) = – 24;2) ( – 7)2 – ( – 4)3 – 34 = 49 – ( – 64) – 81 = 32;3) 13 · 23 – 9 · 23 + 23 = 23 · (13 – 9 + 1) = 8 · 5 = 40;4) 6 · ( – 2)3 – 5 · ( – 2)3 – ( – 2)3 = – 23 · (6 – 5 – 1) = 0 ⋅ ( – 23) = 0.63.1)2)3)4)7 2 ⋅ 715715+ 2=71371353 ⋅ 510 ⋅ 528a ⋅ a ⋅b3c 3 d 5c 9=10 7c d=515+ 4=a9 ⋅ b2713510+3+1=54 ⋅ 515717=a2 +8⋅b3a9 ⋅ b2c1210 2c d=c2d2= 74 ;514519===a10b3a 9b 2c2d251=  ;5551= ab ;.64.1) 1 – 5 =151= 1;03) ( – 10) = 1;2) 4 – 3 =4) ( – 5)134–2=1;64=1=251;2541115)   = 4 =;216236)  7−1=71=2 .3365.1)3)51=   = 4-5 ;45  4 11x71x2)7=   = x – 7;4)12131a93 1 –3 = 21 ;21 = 1a9=   = a – 9.566. 10  3−31) 33=27 −9= 0,027 ; 2) 1000 11 =310153) (0,2) – 4 =  −4=1129=212= (5)4 = 625; 4) (0,5) – 5 =  1;175) – ( – 17) – 1 =−212140=1 ;8180−5= 25 = 32;16) – ( – 13) – 2 = –132= −1.16967.1 1 3+ 4 7;+ ==3 412121) 3 – 1 + ( – 2) – 2 =−333 1 2 ⋅ 27 − 1 535 2==3 ;2)   − 4−2 = 3 − 2 =1616162 4 33) (0,2) – 2 + (0,5) – 5 = 52 + 25 = 25 + 32 = 57; 1  1000 4) ( – 0,1) – 3 – ( – 0,2) – 3 = – −1−1 1 + = – 1000 + 125 = – 875. 125 68.1) 12 – 3 =11232) 210 = 1;< 1;535 5 19 3) (0,6) – 5 =   > 1;−44) 4 19  > 1. 5= 69.11) (x – y) – 2 =3) 3b – 5 c8 =( х − у)3с 8b55) a −1b 2 c −3 =24) 9a3 b – 4 =;b2ac 370.17−31717 1  12)  −  ⋅  −  5  56−4( х + у) 39a 3b46) a 2 b −1c − 4 =;1)   ⋅   =  12) (х + у) – 3 =;−2= 7 2 = 49 ; 1= −  5−3= (−5) 3 = −125 ;;a2bc 4.; 3 10 3) 0,37 ⋅ 0,3−10 = 0,3−3 = 4) 17 −5 ⋅ 17 3 ⋅ 17 = 17 −1 =−3310001 10 ;=  == 3732727 1.1771.11) 97 : 910 = 9−3 =931;729=2) (0,2) 2 : (0,2)−2 = (0.2) 4 = 0.0016 ;2 13 3) 25−1232:  13 254)   :  −10−1=22=  13 4=1322272.( )3) (а )1) а 3−53 7=1691= 42 ;4416.625=54−2( )4) (b )2) b − 2= а −15 ;−47 −4= а − 21 ;= b8 ;= а − 28 .73.()3 = а 3 b −6 = а 6 ;2) а 2 b −1b( )3) 2а(31) аb − 22 −6=2−6a−12=164a12;( )4) 3а3 −4)4 = a 8 b −4 = a 4 ;8b=3−4а−12=181a 12.74. 8b  a1)  7 −2 m −42)  −5n=a −16b−14−3==m12n15b14a16;;2 2 x6 22 x12 y 8 4 x12 y 8=3)  − 4  =;2 3y−5− 4 yx4) 3z3933 −1564 y 3 = − 64 y x= − 9 15 ;9zz x775.() 1−21) x 2 y − 2 − 4 y − 2 ⋅   = ( x 2 − 4) ⋅ y − 2 ⋅ y 2 = x 2 − 4 , yесли х = 5, то x2 = 25 и 25 – 4 = 21;(2 )  a 2 b −1=(a 8 − b 8 )b4⋅)4 a8a4 − b4− a 0 b 4  :=− b424bbb2=(a 4 − b 4 )(a4)(− b4 a4 + b4(b2 ⋅ a4 − b4)если а = 2, b = – 3, то a4 = 16, b4 = 81, b2 = 9 и2⋅ b= a4 − b4)= a4+ b4b2;16 + 81 977== 10 .99976.1) 2000004 = (2 · 105)4 = 24 ⋅ 1020 = 16 · 1020 = 1,6 · 1021;2) 0,00033 = (3 · 10 – 4)3 = 33 · 10 – 12 = 27 ⋅ 10 – 12 = 2,7 · 10 – 11;3) 4000 – 2 = (4 · 103) – 2 = 0,0625 · 10 – 6 = 6.25 · 10 – 8;4) 0,002 – 3 = (2 · 10 – 3) – 3 = 2 – 3 · 109 = 0,125 · 109 = 1,25 · 108.77.1) 0,0000087 = 8,7 · 10 – 6;2) 0,00000005086 = 5,086 · 10 – 8;1= 0,008 = 8 ⋅ 10−3 ;1251= 0,0016 = 1.6 ⋅ 10−3 .4)6253)78, 79, 80.3 · 10– 3мм =3мм = 0,003мм; 0,00000000001с = 10100010 – 4мм = 0,0001мм.81.a8a −7= a8 − 7 + 2 = a 3 ,a−2если а = 0,8, то a3 = 0,512;a15a32)= a15+3−13 = a5 ,a131)5если а =8111, то a5 =   =.3222– 11с;82.1)=−((− 20) ) : ((− 20) ) + 27 −7−6 8(= (− 20)−2−49: (− 20)− 48)+ 41 =1 1 −1 + 5 1+ == ;20 42052)((−17) ) : ((−17) )−4 − 621−  17 −13 −22−2= (− 17) : (− 17) −24262111 1 1−   =  −  −   = 2 − 2 = 0.17 17 17  17   17 83.1) (1,3) – 118⋅ (1,3)127 = (1,3)9 ≈ 10,6;2) (0,87) – 74: (0,87) – 57 = (0,87) – 74 + 57 = (0,87) – 17 ≈ 10,67; 17 3)   19  23 4)   21 −4756 17 :  19  23 ⋅  21 −26−25 17 =  19  23 =  21 −21 19 =  17 21≈ 10,34;31≈ 16,78 .84.1) (786 – 7)4 = 786 – 28 = 5,8 ⋅ 10 – 62;2) (9233) – 6 = 923 – 18 = 4,23 ⋅ 10 – 54;3) (1,76) – 8 ⋅ (35,4) – 8 = (62,3) – 8 = 2,07 ⋅ 10 – 14;4) (0,47) – 5 : (7,81) – 5 = (0,47 : 7,81) – 5 = 1,27 ⋅ 106.85.1) V = (1,54 ⋅ 10 – 4)3 = 3,65 ⋅ 10 – 12 мм3;2) V = (3,18 ⋅ 105)3 = 3,21 ⋅ 1015 км3.86.()(1) a −3 + b −3 ⋅ a − 2 − b − 21  1× 2 − 2 b a=−1(b3) ⋅ (a−111  1⋅ 2 −+ 2abb a)=+ a3 ⋅ a4 b4(2− a −1b −1 + b − 2−1a b ⋅ (b − a )(b + a ) b − ab + a3 3−22)=b3 + a3a 3b 3⋅)1 1=  3 + 3 ×ab−1a 2b 2⋅a 2b 2b 2 − a 2 b 2 − ab + a 2ab( b 3 + a 3 )33( b − a )( a + b )==ab;b−a9()(2) a − 2b − ab− 2 ⋅ a − 2 + a −1b−1 + b− 2a  111 b=  2 − 2 ⋅ 2 ++ 2ab b b  aa=b3 − a3a2b2⋅a2b2b2 + ab + a2=−1)−1==(b − a)(b2 + ab + a2 ) = b − a.b2 + ab + a287.1) 1 = 1;16 = 42 = 4;0 = 0;169 = 132 = 13;2111=   = ;28917 17 2) 3 1 = 1;3330 = 0;3125 = 53 = 5;0,027 = 3 ( 0,3 )3 = 0,3;43)40 = 0;41 = 1;3256 4  4 4=   = ;6255588.1)63)4( )3444111=3 3 = ;27330,064 = 3 ( 0,4 )3 = 0,416 = 24 = 2;443416 4  2 2=   = ;8133 0,0016 = 4 (0,2) 4 = 0,2.= 6 66 = 6 ;( )21264 2 = 12 2 61 1 1  =4  = ;5 25 54)82254 = 8 (152 )4 = 158 = 15 .10 6 = 10 2 = 100 ;2)3312 = 3 4 = 81 ;24889.10= 12 212 = 2 ;2)36 3 = 6 6 21)33)41 24)41 3121 11=  = 3 = ;822316111=  = 4 =.3813 490.1)3− 8 = −2 ;3)3−5)3− 34 3 = −34 ;111= −3=− ;273272)15− 1 = −1 ;4)5− 1024 = − 45 = −4 ;6)7− 8 7 = −8 .591.1) х4 = 81; х = ± 4 81 = ±3 ; х1 = 3; х2 = – 3;2) x5 = −5111 1= 5 −  = − ;; x=5 −32322 23) 5х5 = – 160; х5 = – 32; х = 5 − 32 = – 2.4) 2х6 = 128; х6 = 64; х = ± 6 64 = ±2; х1 = 2 , х2 = – 2.92.1) 6 2 x − 3 — имеет смысл, если2х – 3 ≥ 0 , тогда 2х ≥ 3 , x ≥3,2х ≥ 1,5.Ответ: х ∈ [1.5; + ∞).2)3х + 3 — имеет смысл для любого x.3)32 х 2 − х − 1 — имеет смысл для любого x.2 − 3х ≥ 02 − 3х2 − 3х— имеет смысл, если:≥ 0 , т.е.

2х − 42х − 42 х − 4 > 02222 − 3х ≤ 0 x ≤x ≥x ≥или ; или,поэтому3332 х − 4 < 0 х > 2х < 2x < 24)4Ответ: х ∈ [2; 2).393.1)32)53161 61164 = 3 (−5) 3 + ⋅ 2 6 = −5 + ⋅ 2 = −5 + = −4 ;8884416532 − 0.5 ⋅ 3 − 216 = 2 5 − 3 (−6) 3 = 2 + = 5;22− 125 +113) −4)35)4141181 + 4 625 = − 4 3 4 + 4 5 4 = − ⋅ 3 + 5 = −1 + 5 = 4;333141256 = 3 (−10) 3 − 4 4 4 = −10 − 1 = −11;− 1000 −4450,0001 − 2 ⋅ 0,25 + 5 −= 0,1 − 1 −6) 51 4 1= (0,1) 4 − 2 0,5 2 + 5  −  =32 21= −1,4;21111 3 10 − 9 1+ 3 − 0,001 − 4 0,0016 = − 0,1 − 0,2 = − 0,3 = − == .243333 10303094.9 + 17 ⋅ 9 − 17 = 81 − 17 = 64 = 8;1)22)  3 + 5 − 3 − 5  = 3 + 5 − 2 9 − 5 + 3 − 5 = 6 − 4 = 2 ;23 )  5 + 21 + 5 − 21  = 5 + 21 + 2 25 − 21 + 5 − 21 == 10 + 4 = 14;3+ 23 − 2 ( 3 + 2 )2 − ( 3 − 2 )24)−==3− 23− 23+ 2=3+ 2 6 + 2−3+ 2 6 − 2 2 6 + 2 6== 4 6.3− 2195.1)3(х − 2)3= х − 2 — для любого х.2) т.к.(3 − х )6≥ 0 , то при х<3(3 − х )6и при х≥3(3 − х )6= −(3 − х )3 = ( х − 3)3 .= (3 − х )396.1987 < n < 1988; 19872 < n < 19882 , отсюда3948169 < n < 3952144.Найдем, сколько натуральных чисел между ними3952144 – 3948169 = 3975, а т.к.

n<3952144, то таких чисел 3974.Ответ: 3974 числа.1297.1)3343 ⋅ 0,125 = 3 73 ⋅ (0,5)3 = 3 (7 ⋅ 0,5)3 = 3 (3,5)3 = 3,5;2)3864 ⋅ 216 = 3 33 ⋅ 25 ⋅ 23 ⋅ 33 = 32 ⋅ 2 2 ⋅ 2 2 = 9 ⋅ 43 4 = 36 ⋅ 3 4 ;3)4256 ⋅ 0,0081 = 4 28 ⋅ (0,3) 4 = 22 ⋅ 0,3 = 4 ⋅ 0,3 = 1,2;4)532 ⋅100000 = 5 2 5 ⋅10 5 = 2 ⋅10 = 20 .398.1)3353 ⋅ 73 = 3 (5 ⋅ 7)3 = 353 = 35;2) 4 114 ⋅ 34 = 4 (11 ⋅ 3) 4 = 4 334 = 33;3)5(0,2)5 ⋅ 854)7 711  ⋅ 217 = 7  ⋅ 21 = 77 = 7.33= 5 (0,2 ⋅ 8)5 = 5 1,65 = 1,6;7799.1)332 ⋅ 3 500 = 3 1000 = 103 = 10;2) 3 0,2 ⋅ 3 0,04 = 3 0,008 = 3 0,23 = 0,2;3) 4 324 ⋅ 4 4 = 4 81⋅ 16 = 4 34 ⋅ 24 = 4 64 = 6;4)552 ⋅ 5 16 = 5 32 = 2 5 = 2.100.1)5 102)33 ⋅ 215 = 32 ⋅ 23 = 9 ⋅ 8 = 72;23 ⋅ 56 = 2 ⋅ 52 = 2 ⋅ 25 = 50;821127= 3;3) 4 312 ⋅   = 33 ⋅   = 34)20110 430 ⋅  2 392 1  64= 43 ⋅   == 16.4 2(101 – 102)1)364 ⋅ х 3 ⋅ z 6 = 4 xz 2 ;2)4a 8 ⋅ b 12 = a 2 b 3 ;3)532 ⋅ х 10 ⋅ у 20 = 2 х 2 у 4 ;4)6а 12 b 18 = a 2 b 3 .13102.3334441) 2ab 2 ⋅ 4a 2 b = 2 3 а 3 b 3 = 2ab; 2) 3a 2 b 3 ⋅ 27a 2 b = 3 4 а 4 b 4 = 3ab;3)4ab 4 a 3 c 4 a 4 bc⋅== a; 4)cbbc316ab2116a2=3= .32bab2ab⋅3103.431)364 3 43 3  4 4==   = ;3125555 3)3327 3  3 3=   = ;3 =38822 2)416 4  2 2=   = ;8133 4)519 5 243 5  3 37==   = .323222 53104.1)4324 44= 81 = 34 = 3;4324 : 4 4 = 42) 3 128: 3 2000= 33)33162=3=32 ⋅10364 3  4  4 2=   = = ;1000 10 10 516 33= 8 = 23 = 2; 4)25)()6)( 625 − 5 ):335256585= 5 32 = 25 = 2;2045−= 4 − 9 = 2 − 3 = −1 ;5520 − 45 : 5 =33128625 3 5 33−= 125 − 1 = 53 − 1 = 5 − 1 = 4.555 =3105.14а 6b7 5 5 5= a b = ab;ab 21)5a 6b7 : 5 ab 2 = 52)381x 4 y : 3 3 xy = 33)33x 3 y27 x3 3x:=3= ;22y9xyy34)42ba3:4a8b3=481x 4 y 33= 27 x 3 = 3 3 x 3 = 3 x;3xy16b 4a4=2b.a106.2661)  7 3  = 7 6 = 7; 2)1022108−4−( 32 ) = 324) ( 16 ) = 163)48( 9)−36=9−36=9−12=11921= ;315= 32 5 = 5 32 = 25 = 2;= 16−12=1=116 21.4107.31)3)4)3364 106729 = 3 = 3; 2)9 79 21024 = 2=522= 4 2;9 979 ⋅ 3 = 3 ⋅ 3 = 3 = 3;43625 ⋅ 55 = 12 25 ⋅12 105=12 25 ⋅ 510 =12 125= 5.108.6х31)( х)3)( a ⋅ b) =633=636= х ; 2)  3 у 2  = 3 у 6 = у 3 = у 2 ;26a2126⋅b3= a3b 2 ;24364)  a 2 ⋅ 4 b3  = a 3 ⋅ b 4 = a8b9 ;35) 63216 a b  =  a 6 ⋅ b 6  = a2b; 24134 3 43312126)  27a  = 27 ⋅ a  = 27a3 = 3 (3a)3 = 3a. 109.31)33 3 1 3 3 9 3 33⋅ 2 =⋅ =   = ;242 4222)43 4 3 4 3 27 4  3 3⋅ 6 =⋅=   = ;444 422 41542 4 125 5 4  5 5=⋅ =   = ;58 222 583) 4 15 : 43144) 3 11 : 3 31 3 45 3 3 27 3  3 3=⋅==   = ;34 108222635)  27  = 3 6 = 3; 6) 33616  = 212 = 2 2 = 4.110.1)3ab 2 3 a 5 b 3 a 6 b 3 a 2 b⋅==;ccc2c32)58a3 5 4a 7 5 25 a10 2a 2;⋅==bb2b3b54a 2b 2c ⋅ 4 a 3b3c 23)43abc32a 4b ⋅ 3 4ab4)=432b a 2b 2а 2b 2c ⋅ a 3b3c 2 4 4 4= a b = ab;abc33=51 3 23 a 5b 28a 3 2a a=== ;2b a 2b 22b2b b3535)  a 3  ⋅  b 2  = a 3 b 2 ; 43a3b33= a 2b .6)  4 a3b3  :  ab 2  = ab 2111.49 ⋅ 3 11231)=32504 14 =3   =2 ;5 5533)454 ⋅ 4 12044432249 ⋅ 56 3 7 2 ⋅ 7 ⋅ 8==12553373 ⋅ 232)163=3564= 4 54 ⋅ 24 = 4 27 ⋅ 2 ⋅ 8 ⋅ 3 = 24 ⋅ 34 = 2 ⋅ 3 = 6 ;+ 27 2 −364 = 4326+ 3 − 26 = 4 16 + 3 − 2 = 2 + 1 = 3 ;2313 + 4 18 ⋅ 4 4 −8213= +3− 4 = ;224)27 4 2 ⋅ 32 ⋅ 32 4 4+− 4 =82256 = 335) 3 11 − 57 ⋅ 3 11 + 57 = 3 112 − 57 = 3 121 − 57 = 3 64 = 4 ;6) 4 17 − 33 ⋅ 4 17 + 33 = 4 17 2 − 33 = 4 256 = 4 44 = 4 .112.331)32ab ⋅ 4a 2b ⋅ 3 27b = 23 ⋅ а 3b3 ⋅ 33 = 2 ⋅ 3 ⋅ ba = 6ab ;2)4abc ⋅ 4 a 3b 2c ⋅ 4 b5с 2 = 4 a 4b8c 4 = ab2c ;5a 3 b 2 ⋅ 3a 2 b 33)5544)3ab2 52 xy 2a 5b 5 ⋅ 35348x y ⋅ 4x y45==43ab= 5 a 4b 4 ;16 ⋅ x5 y 6 4= 16 x 4 y 4 = 2 xy .xy 2113.1)3 32) 3) 24)3312a 4  = a 9 + a 6 = a 2 + a 2 = 2a 2 ;33183a18 + x 2  + 2 43a 4b8 − 688x  = x 6 + 2 x 8 = x + 2 x = 3 x ;24 86 12a 3b6  = 2a 4 b 4 − a 6 b 6 = 2ab 2 − ab 2 = ab 2 ;5x6 y12 −  5 xy 2  = 6 x 6 y12 − xy 2 = xy 2 − xy 2 = 0 ;5)  4 x8 y 2= x 4 y − xy 4 ;4  4 2 8 2 8 32 8 4 4 16 8 4 8 −  x y  = x y − x y = ( x y ) − xy 4 = ()5 10(a − 1)5 a= a − 1.6)   5 a 5 a  − 5 a  : a 2 = a5 a − 5 a : 5 a =5a17114.981)7 ⋅ 14 : 3 =2)6,7 ⋅ 23 ⋅ 0,37 = 6,7 ⋅ 23 ⋅ 0,37 = 57,017 ≈ 7,55 ;3)(1,34)−7 ⋅4)(3,44 )−9 : (4,57) −9 = (3,44 ⋅ 4,57) −9 ≈ 3,59 .3≈ 5,72 ;(0,43) − 7 = (1,34 ⋅ 0,43) − 7 ≈ 6,88 ;115.13 ⋅3 91)63( 4−4) ( 9 +3)=633 ⋅ 34 6 6= 3 = 3 ; 2)337 ⋅ 4 343127=373 ⋅ 74112713= 7 12−112= 7;)( 2 + 5 ) = ( 2 ) + ( 5 ) = 2 + 5 = 7 ;4 )( 3 − 2 ) = ( 3 ) − ( 2 ) = 3 − 2 = 1 .3310 + 3 25336 +3333333333333116.4 + 2 3 − 4 − 2 3 = 2 ; (4 + 2 3 )(4 − 2 3 ) = 4 ;2(4 + 2 3 )(4 − 2 3 ) = 2 ;4 + 2 3 − 2 ( 4 + 2 3 )( 4 − 2 3 ) + 4 − 2 3 = 4 ;2 4 + 2 3 − 4 − 2 3  = 22 .