alimov-9-gdz (Алгебра - 9 класс - Алимов), страница 2

Тогда117.1)a− b4−a −4 ba + 4 ab4a +4 b=4+2 3 − 4−2 3 = 2.( a − b )( a + b ) − a ( a + b ) =444444a −4 b444a +4 b= 4 a + 4 b − 4 a = 4 b;2)a −b3a −3 b+a+b3a +3 b3=333(3 a + 3 b ) ( a 2 − 3 ab + a 2 )+3323a+ b333(3 a − 3 b ) ( a 2 + 3 ab + b 2 )3a −3 b= 3 a 2 + 3 ab + 3 b 2 +33+ a − ab + a 2 = 2 а 2 + 2 b 2 = 2 ( a 2 + b 2 );183+3)14a −4 b−14a +4 b⋅( a − b ) = 4a +4 b a− b(()a − 4 b a− b =a − b 4−)4 a +4 b 4 a −4 b  a − b = 4 a − 4 a + 24 b = 24 b ;=− a− ba − b  a+b− 3 ab  : 3 a − 3 b 2 =4) 33 a+ b(( a + b )3=33)3a 2 − 3 ab + b 2  − 3 ab33a+ b33=  a 2 − 3 ab + b 2 − 3 ab  :( a + b)33:( a − b)332=( a − b) = ( a − b) : ( a − b)332323323=118.43x3 = x 2 ;1)2)3a4 = a 3 ;4)5x −1 = x33)5)46b3 = b 4 ;a=1a6;6)7b −3 =−15−3b7;.119.1)1x43) a4= x;−562)6= a −5 ;25y4) b−13= 5 y2 ;3= b −1 ;−215) (2 x ) 2 = 2 x ;6) (3b ) 3 = 3 (3b )−2 .120.112) 27 3 = 3 27 = 3 ;1) 64 2 = 64 = 8 ;234) 814 = 4 813 = 33 = 27 ;3) 8 3 = 3 64 = 4 ;5) 16−344= 16−311= 3 = ;826) 9−32= 9−3 =133=1.2719121.4112151) 2 5 ⋅ 2 5 = 2 5 = 2 3 = 8;213115)( )5124) 4 3 : 4 6 = ;3) 9 3 : 9 6 = 9 6 = 9 2 = 3;2−7 −3 352) 5 7 ⋅ 5 7 = 5; 1 6)  812 2= 7 = 49 ;−4=8−13=1.2122.2246101) 9 5 ⋅ 27 5 = 3 5 ⋅ 3 5 = 3 5 = 32 = 9;2)2732⋅ 49 33)3144 43: 944)315023: 62=2734⋅73=6733= 7 2 = 49;3 144  4= = 16 4 = 23 = 8;933 150  2= = 25 2 = 53 = 125.6123.341 −1 −1)   4 +   3 = 2 3 + 2 4 = 8 + 16 = 24 ; 16 82) (0,04)−32− (0,125)= 125 − 4 = 121;926−23 1 =  25 −321− 8−2332= 25 2 − 8 3 = 5 3 − 2 2 =43) 8 7 : 8 7 − 3 5 ⋅ 3 5 = 8 − 3 2 = 8 − 9 = −1 ;4) (5−25 ) −531+ ((0,2 ) 4 ) − 4 = 5 2 +  5−3= 25 + 125 = 150 .124.661) 3 a ⋅ 6 a = a 2 ⋅ 6 a = a 3 = a , при a=0,09,2)2066b : 6 b = b3 : 6 b = b 2 = 3 b , при b = 27,a = 0,09 = 0,3;b =327 = 3 ;3b ⋅ b23)4)63b66b3 ⋅ b 4=66= b6 = b = 1,3;ba ⋅ 4 a ⋅ 12 a 5 = 12 a 4 ⋅ 12 a3 ⋅ 12 a5 = 12 a12 = a = 2,7.125.11 1+251) a 3 ⋅ a = a 32)3)1b21⋅ b3 ⋅ 6b=31: b61 1−b3 6b=4= a6 ;5 1+b6 6=4 1−34) a 3 : 3 a = a 3=6b62 1−b6 6=b;1= b6 ;=a;5) x1,7 ⋅ x 2,8 : x5 = x 4,5 : x 2,5 = x 4,5− 2,5 = x 2 ;6) y − 3 ,8 : y − 2 , 3 ⋅y3 = y− 3 ,8 + 2 , 3 +32= y0 =1.126.1) 2 2−35⋅832:9:  42) 31+ 233) 61+ 234)  51+2= 2 2−355 +3 53= 22 = 4 ;3= 31+ 2 2 − 2 2 = 3 ;3⋅ 9 3  = 61+ 2 3 : 6221− 2= 51− 2 = 5 −1 =3= 61+ 23 −2 3= 6;1.5127.1) (а 4 )−34⋅ (b−23 ) −6= a −3 ⋅ b 4 ;1  6 4 12 a2)   −3   = a 24b12  b  (103)  x 0, 4 ⋅ y1, 2 4) x − 22112)= a 2b ;(= x 0 , 2 ⋅ y 0, 6 1 ⋅  x − 2 −1 )10= x2 ⋅ y6 ;2 +1= x−22⋅ x2+ 22 +1= x−22 + 3+ 2 2= x3 .21128.4−124 1−4+2а 3 (а 3 + а 3 )а3 3 +а3 3а + а2а (а + 1)1)= 1 3===а;1 11  31 +аа +11+−−а 4 а 4 + а 4  а 4 4 + а 4 41b52)⋅  5 b 4 − 5 b −1 2b 3 3=3 −2  b− b 3)4)5а331−⋅ b −1 − ab 33a2 − b21a3b61+ b3=1 4+51 1−5b5−b52 1+b3 32 2−−b3 32 2ab −1  a 3 − b 32a32−b31 1 1 1−a 3b 3  b 2 3aa +6 b=1a6+=b −1= 1;b −1=а;b1 1−a2 31+ b61 1 11 a 3b 3  b 6 + a 6 1 1= = a 3b 3 .11a6 + b6129.5111 5 −1− −−333333⋅ 6 = 2 ⋅ 3 3 22 − 32 ⋅ 3 2 ⋅ 3 3 =1) 2 ⋅ 3 − 3 ⋅ 24−9 3 3= 3 3 ⋅ 2 ⋅ 3 = −5;2⋅ 3(313 12)  5 4 : 2 4 − 2 4 : 5 43)  22234)  (0,5)5 22+  3−5−3 +1 (1 1 54 24 ⋅ 4 1000 = − 33 24 545−4 − 0,3 3( ))3 −1)35−24 = 3; ⋅ 4 10 3 =⋅10310 4= 22 + 33−1 = 22 + 32 = 4 + 9 = 13;−311=   − 42 = 8 − 2 = 6 .2130.111)1a932) =1a21 211  4 66 = a9 a3  = a9+9 = a3 ;⋅ 6 a3 a =  1112 1 −21−− 6ab − 2 + (ab )− 6  ⋅ ab 4 =  a 3 b 3 + a 6 b 6 a 6 ⋅ b 3 =11a 9  aa 31+b2= a + b;13)4)1b 12 ⋅ 3b4 b =( a + b ) a33=( a) +( b)3333231 1b 12  bb 411511  5 33 = b 12  b 4  = b 12 ⋅ b 12 = b 2 = b ;  2+ b 3 − 3 ab  =( a + b ) ( a ) + ( b )333232− 3 a 3 b  == a + b.131.1)2)x− y1x211  11 1 x 2 + y 2  x 2 − y 2 11 = x2 − y2 ;=11+ y2x2 + y21  11 1 a 4 − b 4  a 4 + b 4 11a− b  = a4 + b4 ;=1111a4 − b43)1m21+ n2m + 2 mn + na4 − b41=1m2 + n2 1m21+ n22=11m21+ n2;2 1  c 2 − 1112c − 2c + 1  = c 2 −1 .4)=1c −1c 2 −123132.11b b   21) 1 − 2: a −b2+ aa ( a − b)a⋅( a − b)2=1  1ab2)  a 3 + b 3  :  2 + 3 + 3  = ba  ( a + b)=( a + b)3=ab33313)9a4 − a41a4323−5− a4b−3124)312ba− b−3−=( a − b)261a412a+ b=3142(1 − a )a +aa −b3−3a 2 + 23 ab + 3 b 23ab;a2 − a−( a + b ):(1 − a )a − (1 − b )b=− b21b21 − a −1ba −b=−aba +3 b+b= 1 + a − (1 − b) = a + b;a −a11;a=222 = 1 − b  ⋅a −1313bb2(1 + b)b(a − b )a=−12a− b−−1212=(a − b )a−13a+ b3a(a − b )(−=a) ( a − b )(a − b) =a+ b−a−ba −b= a + b − a + b = 2 b.133.1)−3a2a+ b−1ab 2b− a22a − 4ab( a + b )( a − b )=24a23 1− a 2b 23+ a2=3a2−22a − 4ab=a−b3a2a+ b1+ ab 2+1b2aa− b( a − b ) ( a + b )− 2a( a + b )( a − b )b + ab − 2a 2 + 4ab 5ab − a 2;=a −ba −b2−+ 4ab==2)=3xy − y 2−x− yy yx− y3 xy − y 2(x− y)(y x−x+ y=x+ y)y x−x+ yy y−x− y3 xy − y 2 − y y(=–)x+ y −y xx− y(x− y)=333xy − y 2 − y 2 x − y 2 − yx + y 2 x 2 xy − 2 y 2 2 y (x − y )==== 2 y;x− yx− yx− y13)33a +3 b−2a33a+ b2+b3− 3 ab3=2a33− ab32+b3(− 3 a +3 ba+b)( a + b ) =333a 2 − 3 ab + b 2 − a 2 − 23 ab − b 2 − 33 ab;=a+ba+b=34)3a2 − 3 b2a −3 b( a − b )⋅ 3−3333a−b−22a 3 + 3 ab + b 3a + 3 ab + 3 b 2 33a + ab + b2=( a − b )( a − b ) −3333a −3 b= 3 a + 3 b − 3 a + 3 b = 23 b .134.1)(a − b )3a −3 b−(a+b113) 3 a 2 + 3 ab + 3 b 2  3 a − 3 b=−33a− ba3 +b3 3 a 2 − 3 ab + 3 b 2  3 a + 3 b−=3a +3 b(333)3= a 2 + 3 ab + b 2 − a 2 + 3 ab − b 2 = 23 ab;252)a+b2a31 1− a 3b 32a32+ b3−2+ b3a −b2a31 121 12  11 2 a 3 − a 3b 3 + b 3  a 3 + b 3 −=21 12a 3 − a 3b 3 + b 3+ a 3b 3 + b 31 12  11 2 a 3 + a 3b 3 + b 3  a 3 − b 3 11  111 − = a 3 + b 3 −  a 3 − b 3  = 2b 3 ;−21 12a 3 + a 3b 3 + b 33)−a −b122  21 12a 3 + b 3 −  a 3 + a 3b 3 + b 3  3 = ab ;=a −bb−a11a31− b311a3 −b34)+a+b211111a 3 − b 3 + a 3 + b 3 23 a.==a+ba+b12a 3 − a 3b 3 + b 3135.1) 3 3 + 3 4 ≈ 3,02; 2) 3 7 + 5 10 ≈ 2,04; 3) 54)( 2)333≈ 16,24;≈ 1,49; 5) ππ ≈ 36,46.136.111) 2 3 < 3 3 ;3) 53<732) 5−454) 21−;<32−45, т.к.> 31−2,1554т.к.<15121;342>131137.1) (0,88) 52)   12 2616−1411 6 6>  , 11 < (0,41)−1886 88  6  6  6> ,ит.к. >  ;100 11  100  11 141,112 100  12  4  100  4т.к.и   << ;541 5 41 2.3) (4,09 )3 11 4)   12  3 < 4  25 2− 5 12 >  13 323 , т.к.

 4,09 < 4  ;25 − 5,т.к.12 13 12 >и  11 12 11 5 13 >  12 5.138.152xx=6 .1Тогда 2 x = .51.Отсюда x =101) 63) 71− 3 x2) 3 = 27 ;xх = 3.10=7 .4) 22+ x2+ x2 x +12 x +1= 32 ,52=2 .Тогда 2х + 1 = 5, х = 2.Поэтому 1 – 3х = 10,х = – 3.5) 433 =3 ;16)  5=1;04 x −3=5,53− 4 x = 5 ,4=4 .Поэтому 2 + х = 0,3 – 4х = 1,1x= .2х = – 2.139.1)722222 3− 2 171 1 −  =7 =  ; 2 3 6 62 1 71 1 4−37  −  =7  = 3 4 12  12 т.к.1 12> , а > 0,6 127то7221 11 1 −  >7 −  .233 4272)51 11 − 1 5 433и51 11 − 1  ;7 63331 1 5 1 25 − 24 1 − 1  = 5  =  ;4520 20 53331 1 49 − 48  1 51 − 1  = 5  =  ;7 6 42  42 5т.к.113>, а> 0,20 425то53311 1 11 − 1  > 5 1 − 1  .57 4 6140.1) 32) 32− y= 27 , 35− 2 x=1; 33)1x −1924)13− y27 32− y5− 2 x−3 = 0 ;3= 3 .

Тогда 2 – у = 3 и у = – 1.0= 3 . Поэтому 5 – 2х = 0 и х = 2,5.1x −192− 81 = 0 ;=3; 1 3 3− y 3 3 12 x −123= 3 . Тогда х – 2 = 1 и х = 3.= 3 4 . Тогда 9 – у = 4 и у = 5.141.11)  92 x −5=35 x −8−2;  3 2 x −5=35 x −8;3−4 x +10 = 35x −8 .Тогда 10 – 4х = 5х – 8,9х = 18 и х = 2.x−41=   ; 2 4 x −9 = 2 − x + 4 .2Поэтому 4х – 9 = – х + 4,5х = 13 и х = 2,6.xx +1313) 8 ⋅ 4= ;1623x ⋅ 22 x + 26 = 2−4 .Тогда 3х + 2х + 26 = – 4, 5x = – 30; х = – 6.2) 2284 x −925 x − 24)1= 552 x −4−x − 7 ,5;1− +2 = 5 x 7 ,5 .5Тогда 2х – 4,5 = – х + 7,5,3х = 12 и х = 4.142. 1 1)  3(33−2 x +112 ) 2 x +1− x−12=3( )x= 3 3 ,32)x= 32 ,3x2Тогда − x −23) 9(3 )2 3х+ 4x −132x,4x=23 .х −1 4= x,33.Поэтому1 3= x,2 2х – 1 = 4х,– 2,5х = 0,51и x=− .53x+4( 2) 2 = 3 2x −133х = – 1и x=−⋅ 3=27x −13( )⋅ 3 = 33,х −14)2,1.38( 2)х31х22=43 х−22,1= 2 2(3 х− 2)⋅22 .11х = 2 (3 х − 2 ) + ,22116 х=622и х = 1.36х + 8 + 1 = 33х – 3.Тогда 3 −Тогда 6х + 9 = 3х – 3,3х = – 12 и х = – 4.143.21) log 7 49 = log 7 7 = 2 ;13) log 1 4 = log 1  222 62) log 2 64 = log 2 2 = 6 ;−2= −2 ;4) log 3−31= log 3 3 = −3 .2729144.1) lg23 ≈ 1,4; 2) lg131 ≈ 2,1; 3) 40lg2 ≈ 12; 4) 57lg3 ≈ 27,2.146.1+ lg 7, x ≈ 0,92;21) 102x – 1 = 7, 2x – 1 = lg7, x =2) 101 – 3x = 6, 1 – 3x = lg6,1+ lg 6x=, x ≈ 0,07.3146.422625 100  25 1) (0,175)0 + (0,36)− 2 − 1 3 = 1 + ; −1 =   =36981 113 1000  32) 1−0,43 − (0,008) + (15,1) = 1 −  +1 = 8 = 2−3010310= 2 − = −3 ;2231−22125 14 1 35+4=− +4=3)   −   + 4 ⋅ 379 0 =   − 3527427163   25 11 25111;=+ ==516 348484) (0,125)=−1323+   − (1,85)0 =4130,125+919−1 =+ −1 =160,5 1699+ 2 −1 = 1 .1616147.()1) 9,3 ⋅ 10−6 : 3,1 ⋅ 10−5 =−69,3 ⋅ 10−63,1 ⋅10−5= 3 ⋅ 10−1 = 0,3 ;72) 1,7 ⋅ 10 ⋅ 3 ⋅ 10 = 5,1 ⋅ 10 = 51 ;3) 8,1 ⋅ 1016 ⋅ 2 ⋅ 10−14 = 16,2 ⋅102 = 1620 ;()4) 6,4 ⋅10 5 : 1,6 ⋅10 7 =306,4 ⋅10 51,6 ⋅107=410 2= 0,04 ;−1−231 1 1  15) 2 ⋅10 −1 +  6 0 −  ⋅   ⋅   ⋅  − 6  3  3  41 2 ⋅ (−4) ⋅ 3 1 87= += − = − = 1,4;55⋅35 5516) 3 ⋅10 −1 −  8 0 − 83 2 3−4=− == −0,1.10 510−11⋅ 4−341 5⋅  ⋅ 4 7−1=−1=1 6 32+ ⋅ ⋅ (− 4 ) =5 5 333 8 1 7− ⋅ ⋅ =10 7 4 5148.−2 1 5 2 5 х3 ⋅ х6  х6 ⋅ х6 1)  1  =  1  х6  х6 7 18132==1 ;при x =9 x 2 4949 2 1 а3 ⋅ а92)  − 2 а9−3−2 6 1 а9 ⋅ а9=  −2 а9при a = 0,1, a3 = 0,001,1a3 7 х6 = 1   х6  −2= х−2 =−3 7 2= а9 ⋅ а9 1х2,−3= (а )−3 =1,а3= 1000.149.( 125х − 8х )− ( 27х − 64х ) = (5 х − 2 х )− (3 х − 4 х ) = 42) ( х + 16 х )+ ( 81 х − 625 х ) =1)334=44334х + 2 4 х + 34 х − 5 4 х =33333х;44х; 3 + 1 − а 2  1 + а2 3 3 + 1− а=+ 1− а  :=1;3)  1+ а1+ а1 + а  3 + 1 − а 2 х2 − у2 − хх1 :  х2 − у2 − х  =.4) 1 −=2222 2222−−хухухухух−−−31150.1) 75 х −1= 49 ; 75 х −12=7 .Тогда 5х – 1 = 2; 5х = 3 и х =3.52) (0,2)1− х = 0,04 ; (0,2)1− х = (0,2)2 .Поэтому 1 – х = 2 и х = – 1.13)  73 х +32х=7; 7−3 х − 3=72х.Значит, – 3х – 3 = 2х; – 5х = 3 и х = –3.52х−2 х5 х −71=3 .=  ; 33Отсюда, 5х – 7 = – 2х; 7х = 7 и х = 1.4) 35 х −7Проверь себя1.31) 32)3)= 5+5−5:3−7−2310 ⋅ 32 −325 2⋅ 25−1−233+  2  −1 222733⋅ 2 +    = 3 − 2 += 9−4+3 =8 ; 3  888448= 3 2 ⋅ 2 − 3 8 = 18 − 2 = 16 ;32⋅ 3253 3( ):532− 48 32:63= 25 + 5−12−83=1− 4 = 1,2.52.8600 = 8,6 ⋅ 103;0.0078 = 7,8 ⋅ 10 – 3;1) 8,6 ⋅ 103 ⋅ 7,8 ⋅ 10 – 3 = 67,08;2) 8,6 ⋅ 103 : 7,8 ⋅ 10 – 3 =643⋅10 .393.1)32(3 х −9 ⋅ 2 х5= 6 ; 2) х−1 + у−1х−4−2)⋅  ху1 =у+х⋅ (ху)2 = (х + у)ху .ху4.55а333= а3 ⋅ а−23⋅a−34= а⋅а−341−=а3411= а 4 ; при а = 81, то a 4 = 3 .а2 ⋅ а 45.22а) (0,78) 3 > (0,67 ) 3 , т.к.