alimov-7-algebra-gdz (Алгебра - 7 класс - Алимов), страница 4

Пусть х ч. – время, необходимое туристам для преодоления оставшегося расстояния. Составим уравнение:⎛3,5 ⋅ ( x + 1) = 5 ⋅ ⎜ x −⎝1⎞⎟ ; 3,5x + 3,5 = 5x − 2 ,52⎠15, x = 6 ⇒ x = 4 ; 3,5 ⋅ (4 + 1) + 3,5 = 21 км – прошли туристы.Ответ: 21 км127. Пусть равнинный участок – х км, тогда остальной — (9 – х) км,составим уравнение299 − x 9 − x 2x41++=3;46560135 − 15x + 24 x + 90 − 10 y = 221y = 225 − 221 ; y = 4 км.Ответ: 4 км128. 100% – 84% = 16% – сушеные яблоки16 : 0,16 = 100 (кг) – свежие яблоки.Ответ: 100 (кг)129. 100% – 12% = 88% – кофе готовый к употреблению4,4 : 0,88 = 440 : 88 = 5 (кг) – свежий кофе.Ответ: 5 кг130. 1) 173x + 199 ,6 = 2517 ,8 ;2) 24 ,8 x + 25,47 = 71,35x = (2517 ,8 − 199 ,6) :173 ;x = (71,35 − 25,47) : 24 ,8x = 13,4 ;x = 185,131. 1) 2 x − 1 = 3 ;2) 1 − 5x = 2а) −(2 x − 1) = 3 ;а) −(1 − 5x ) = 2−2 x + 1 = 3 ;б) 2 x − 1 = 3 ;x2 = 2 ;−1 + 5 x = 23x1 =5б) 1 − 5x = 2−5x = 13) x − 1 = x + 3 ;154) 2 x − 1 = x − 1x1 = −1 ;x2 = −а) x − 1 = x + 3 ;−1 = 3 – решений нет.б) x − 1 + − x − 3 ;x = −1 ;132.30а) 2 x − 1 = x − 1x1 = 0б) 2 x − 1 = 1 − x2x2 =375= 25 (м/с) скорость сближения поездов;30,025= 0,025 ⋅ 3600 = 90 км/ч25 м/с =1: 360090 − 40 = 50 (км/ч) – скорость встречного поезда.Ответ: 50 км/ч.Глава III.

Одночлены и многочлены§ 9. Степень с натуральным показателем133. 1) a = 5 см.2) a =s = 52 = 25 (см2)s=1м.21 2(м )414) a = 2 ,7 дм.41 1 13 139s = 3 ⋅ 3 = ⋅ = 10 км s = 2 ,7 ⋅ 2 ,7 = 7 ,29 (дм2)4 4 4 4163) a = 3 км134. 1) a = 2 м.v = 23 = 8 (м3)3) a =2) a = 3 дм.v = 33 = 27 (дм3)1км54) a = 0,4 м.31⎛ 1⎞v=⎜ ⎟ =(км3)⎝ 5⎠125v = (0,4) = 0,064 (м3)3135. 1) 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 262)1 1 1 1 1 ⎛ 1⎞⋅ ⋅ ⋅ ⋅ =⎜ ⎟3 3 3 3 3 ⎝ 3⎠4) m ⋅ m ⋅ m ⋅ m ⋅ m = m555) ( x − y ) ⋅ ( x − y ) ⋅ ( x − y ) = ( x − y )3) x ⋅ x ⋅ x ⋅ x = x 4136.

1) 5 ⋅ 5 ⋅ 8 ⋅ 8 ⋅ 2 ⋅ 2 = 52 ⋅ 82 ⋅ 226)51 1 1 12 ⎛ 1⎞= (0,3) ⋅ ⎜ ⎟⎝ 7⎠7 7 7 743) 0,3 ⋅ 0,3 ⋅ ⋅ ⋅ ⋅2) 6 ⋅ 6 ⋅ 7 ⋅ 7 ⋅ 3 ⋅ 3 ⋅ 3 = 62 ⋅ 72 ⋅ 33 4)137. 1) 9 ⋅ 9 ⋅ 9 ⋅ a ⋅ a ⋅ a = 93 ⋅ a 3m m m m m ⎛ m⎞⋅ ⋅ ⋅ ⋅ =⎜ ⎟n n n n n ⎝ n⎠332 2 2⎛ 2⎞2⋅ ⋅ ⋅ 2 ,3 ⋅ 2 ,3 = ⎜ ⎟ ⋅ (2 ,3)⎝ 3⎠3 3 32) x ⋅ x ⋅ x ⋅ x ⋅ 3 ⋅ 3 = x 4 ⋅ 323133)⎛x⎞x x x⋅ ⋅ ⋅ (x − y ) ⋅ (x − y ) = ⎜⎜ ⎟⎟ ⋅ (x − y )2y y y⎝ y⎠4)a a⎛a⎞⋅ ⋅ (8a − b ) ⋅ (8a − b )(8a − b ) = ⎜ ⎟ ⋅ (8a − b )3b b⎝b⎠2138.

1) 31⋅432⋅ .....⋅ 3 ⋅ x1⋅4x2⋅ .....⋅ x = 321 ⋅ x12434321 раз12 раз52⋅ .....⋅ 5 ⋅ b1⋅4b2⋅ .....⋅ b = 516 ⋅ b312) 51⋅4434316 раз31 раз72⋅ .....⋅ 7 ⋅ p ⋅ p ⋅ ..... ⋅ p = 7 n ⋅ p153) 71⋅44314243n раз15 раз62⋅ .....⋅ 6 ⋅ a1⋅4a2⋅ .....⋅ a = 613 ⋅ a k4) 61⋅4434313 разk раз3) a ⋅ a + a ⋅ a + a ⋅ a = a 2 + a 2 + a 2 = 3a 2139. 1) p ⋅ p ⋅ p + q ⋅ q = p3 + q 22) a ⋅ a + b ⋅ b ⋅ b ⋅ b = a 2 + b4 4) x ⋅ x ⋅ x + x ⋅ x ⋅ x = x 3 + x 3 = 2 x 3140. 1) 113 = 11 ⋅ 11 ⋅ 112) ( −1,25) = (1,25) ⋅ (1,25) ⋅ (1,25) ⋅ (1,25)43) (2a ) = 2a ⋅ 2a ⋅ 2a ⋅ 2a ⋅ 2a54) (a + b) = (a + b) ⋅ (a + b) ⋅ (a + b) ⋅ (a + b)4141.

1) 23 = 8;2) 32 = 9;3) 104 = 10000; 4) 53 = 125142. 1) 15 =1;2) (– 1)7 = – 1;3) 015 = 0;143. 1) ( −5) = −125 ;3⎛⎝1⎞4⎠⎛ 2⎞⎝ 3⎠323) ⎜ −2 ⎟ =144. 1) ⎜ ⎟ =⎛ 2⎞⎝ 7⎠811=5 ;16168;2723) ⎜ 1 ⎟ =8132=1 ;4949145. 1) 2 ⋅ ( −3) = 2 ⋅ 9 = 18 ;2324) 05 = 02) −53 = −125⎛ 1⎞⎝ 4⎠24) −⎜ 2 ⎟ = −5⎛ 3⎞⎝ 5⎠22) ⎜ ⎟ =⎛ 1⎞⎝ 3⎠1169253⎛ 7⎞⎝ 3⎠34) ⎜ 2 ⎟ = ⎜ ⎟ =34319= 1227272) −5 ⋅ ( −2) = 5 ⋅ 8 = 40312123) − ⋅ ( −4) = − ⋅ 16 = −8 ;223234) − ⋅ ( −3) = − ⋅ 9 = −62146. 1) 12 ⋅ 102 − 53 ⋅ 10 = 1200 − 1250 = −502) 92 ⋅ 2 + 200 ⋅ (0,1) = 81 ⋅ 2 + 200 ⋅ 0,01 = 162 + 2 = 1642⎛ 1⎞⎝ 3⎠43) ⎜ ⎟ ⋅ 27 + (0,1) ⋅ 50000 =5⎛ 1⎞⎝ 4⎠34) 103 : 40 − ⎜ ⎟ ⋅ 128 =115⋅ 27 + 0,00001 ⋅ 50000 = + 0,5 =81361000 128−= 25 − 2 = 23406412743 = 1 ⋅10 4 + 2 ⋅10 3 + 7 ⋅10 2 + 4 ⋅101 + 35043201 = 5 ⋅ 106 + 4 ⋅ 104 + 3 ⋅ 103 + 2 ⋅ 102 + 113027030 = 1 ⋅10 7 + 3 ⋅10 6 + 2 ⋅10 4 + 7 ⋅10 3 + 3 ⋅10147.

1)2)3)4)12350107 = 1 ⋅10 7 + 2 ⋅10 6 + 3 ⋅10 5 + 5 ⋅10 5 + 1 ⋅10 2 + 7148. 1)2)3)4)2 ⋅ 105 + 3 ⋅ 104 + 5 ⋅ 103 + 1 ⋅ 102 + 2 ⋅ 10 + 1 = 2351213 ⋅ 106 + 5 ⋅ 105 + 3 ⋅ 104 + 2 ⋅ 103 + 3 ⋅ 10 + 7 = 35320377 ⋅ 105 + 1 ⋅ 103 + 5 ⋅ 102 + 8 = 7015081 ⋅ 105 + 1 ⋅ 103 + 1 = 101001149. 1) 2 ⋅ 104 + 3 ⋅ 102 + 6 = 20306 – не делится на 5, т.к.

последняяцифра ни 0 ни 5; 2 + 3 + 6 = 11 на 3 не делится.2) 4 ⋅ 105 + 3 ⋅ 104 + 2 ⋅ 10 + 5 = 430025 – делится на 5, т.к. оканчивается цифрой 5, а на 3 не делится, т.к. (4 + 3 + 2 + 5) – не делится на 3.3) 7 ⋅ 103 + 8 ⋅ 102 = 7800 – делится на 5, т.к. оканчивается цифрой0; делится на 3, т.к. (7 + 8) = 15; 15 : 3 = 5.4) 5 ⋅ 104 + 3 ⋅ 103 + 10 = 53010 – делится на 5, т.к. оканчивается на0; делится на 3, т.к. (5 + 3 + 1) = 9; 9 : 3 = 3.150. 1) 249 = 2 ,49 ⋅ 102 ;2) 781 = 7 ,81 ⋅ 1023) 84340 = 8,434 ⋅ 104 ;4) 80005 = 8,0005 ⋅ 1045) 3100,2 = 31002,⋅ 103 ;6) 127 ,48 = 1,2748 ⋅ 102151.

Sп.п.к. = 6k2 см2;Vm = k3 см3.33152. 1) m2 ;3) (c + 3) ;22) a 3 ;⎛ 1⎞⎝ 2⎠2⎛ 1⎞⎝ 2⎠44) c2 + 321 1>4 16153. 1) ⎜ − ⎟ > ⎜ − ⎟ , т.к.2) 23 < 32 т.к. 8 < 93) ( −0,2) < ( −0,2) т.к. −0,008 < 0,043⎛ 1⎞⎝ 2⎠32⎛ 1⎞⎝ 2⎠298>72 724) ⎜ ⎟ > ⎜ ⎟ т.к.154. 1) 3x + ( −0,1) = ( −0,485) ;32) ( −1,415) + 2 x = ( −9 ,15)423x = ( −0,485) + 0,13 ;32 x = ( −9 ,15) − 1,415243x>0;x<03) ( −7 ,381) − (1 − x ) = (8,0485) ; 4) (10,381) = ( −0,012) − 2 x32(7 ,381)3 − 1 + x = (8,0485)2 ;2x = (8,0485) + 1 + 7 ,381 ;352 x = −(0,012) − (10,381)53x<0x>0155.

1) 27000000000000000000 = 2,7 ⋅10192) 30800000000000 = 3,08 ⋅10133) 1000000 = 106156. 510млн.км2= 51, ⋅ 108 км21000млрд.км = 1012 км157. 1л. = 1дм3 в 1 дм3 – 0,00001 мг золота1км3 = 1012 дм3 в 1012 дм3 – х мгПолучаем пропорцию:10,00001=x = 1012 ⋅ 0,00001x1012x = 107 (мг.)107 мг = 10 кгОтвет: в 1 км3 морской воды содержится 10 кг золота.3⎛ 3⎞ ⎛⎝ 7⎠ ⎝1⎞3⎠158. 1) ( −18, ) ; ⎜ ⎟ ; ⎜ −1 ⎟ ;234⎛ 1⎞⎝ 7⎠32) ( −7) ;( −0,4) ; ⎜ ⎟ ;( −15,) .332159.

Сложим цифры на которые оканчиваются данные степени:1) 33 + 43 +53 = ..... 7 + ..... 4 + ..... 5 = 6 – 6 последняя цифра2) 313 + 1013 + 1813 = .....3+ .....0+ .....8 = .....1 – 1 последняя цифра3) 214 + 344 + 464 = .....1+ .....6+ .....6 = .....3 – 3 последняя цифра4) 155 + 265 + 395 = .....5+ .....6+ .....9 = .....0 – 0 последняя цифра§ 10.

Свойства степени с натуральным показателем160. 1) c3 ⋅ c2 = c5 ;⎛1 ⎞⎝2 ⎠72) a 3 ⋅ a 4 = a 7⎛1 ⎞⎝2 ⎠⎛1 ⎞⎝2 ⎠84) (3b) ⋅ (3b) = (3b)63) ⎜ a⎟ ⋅ ⎜ a⎟ = ⎜ a⎟ ;7(опечатка в ответе задачника).161. 1) 23 ⋅ 22 ⋅ 24 = 29 ;2) 32 ⋅ 35 ⋅ 33 = 3103) ( −5) ⋅ ( −5) ⋅ ( −5) = ( −5) ;6344) ( −6) ⋅ ( −6) ⋅ ( −6) = ( −6)133162.

1) (− 2,5)3 ⋅ (− 2,5)8 = (− 2,5)112) (−3) ( x − a ) ⋅ ( x − a ) = ( x − a )7163. 1) 32 = 25;4) 256=28;102756)5⎛ 9⎞⎝ 7⎠3167. 1) ⎜ − ⎟ : ⎜ − ⎟ = ⎜ − ⎟ ;3) x 21 : x 7 = x14 ;203) 1024=2106) 32 ⋅ 64 = 25 ⋅ 26 = 2113) 8 : 22 = 26)210= 2923) 729 = 366) 243 ⋅ 27 = 35 ⋅ 33 = 382) 27 : 32 = 33 : 32 = 34) 81 : 9 = 34 : 32 = 32315= 314 ;3⎛ 9⎞ ⎛ 9⎞⎝ 7⎠ ⎝ 7⎠5= 22 ;2) 27 = 33;5) 36 ⋅ 81 = 36 ⋅ 34 =310;166.

1) 34 : 9 = 34 : 32 = 32;3) 243 : 27 = 35 : 33 = 32;8152) 32 : 23 = 25 : 23 = 22;2125x 5 5x 75x) (− ) = (− )126662) 128=27;5) 25 ⋅ 128 = 25 ⋅ 27 = 212 ;4) 256 : 32 = 28 : 25= 23; 5)5)74) (n + m) ⋅ (n + m) = (n + m)17164. 1) 64 : 4 = 16 = 24;165. 1) 81 = 34;4) 243 = 35;23834= 3418171⎛ 1⎞ ⎛ 1⎞2) ⎜ ⎟ : ⎜ ⎟ =17⎝ 17 ⎠ ⎝ 17 ⎠4) d 24 : d 12 = d 1235624⎛ 3y ⎞ ⎛ 3y ⎞⎛ 3y ⎞⎟ :⎜ − ⎟ = ⎜ ⎟ ;⎝ 4⎠ ⎝ 4⎠⎝ 4⎠168. 1) ⎜ −3) (a − b) :(a − b) = (a − b) ;7169.

1)3)170. 1)3)2 ⋅ 3352= 2⋅3 = 6 ;233 5 ⋅ 31036 ⋅ 378 ⋅ 33315=3132 4 ⋅ 26 ⋅ 2 3=9;213=25 ⋅ 2 72= 2⋅3 = 622 ⋅ 358 ⋅ 5 75 4 ⋅ 59113 ⋅ 42112 ⋅ 436 ⋅ 334)35 ⋅ 3 ⋅ 3=515= 25513= 11 ⋅ 4 = 44=39= 32 = 9372) x : 24 = 22 ;3) x ⋅ 2 6 = 2 8x = 33 ⋅ 32 = 35 ;x = 243 ;x = 22 ⋅ 24 = 2 6 ;x = 64 ;4) x ⋅ 35 = 38 ;5) 55 ⋅ x = 57 ;x = 28 : 2 6 = 2 2x=46) 46 ⋅ x = 48x = 38 : 35 = 3 3 ;x = 27 ;x = 57 : 55 = 52 ;x = 48 : 4 6 = 4 2x = 25 ;x = 16 .( ) =a ;2) ( a ) = a3) ( a ) ⋅ a = a ⋅ a5) a ⋅ a ⋅ ( a ) = a6308 756172. 1) a 52 578102 45812= a18 ;⋅ a 8 = a 20 ; 6) a 3543512236 43 5241593 5a12415a4123( ) = a ⋅a = a⋅ (a ) ⋅ a = a ⋅ a353 334) a5 ⋅ a 2( ) ( ) = a :a = a ;2) ( a ) :( a ) = a : a = a ;(a ) ⋅ a = a ⋅ a = a ⋅ a = a3)173. 1) a 7 : a 33623 ⋅ 322)171.

1) x : 32 = 33 ;34) (m + n )10 : (m + n )5 = (m + n )54)=2;21252)= 4 ⋅ 3 = 12 ;2 ⋅ 322) (2a ) :(2a ) = (2a )47;61169= a 154)( )(a )a8 ⋅ a 443 4(c ) ⋅ c174. 1)(c )2 3a128d3 ⋅ d52 3=a 24a12c14c12= a12 .= c2 при с = – 3(– 3)2 = 9;2=(d )c12=4⎛ 2⎞;⎜ ⎟ =⎝ 7⎠4927при c =c6 ⋅ c8=3 42)a 8 ⋅ a16=d8d6= d 2 при d =21⎛ 1⎞;⎜ ⎟ =⎝ 4⎠1614( −10)2 = 100 .при d = −10175. 1) 220 = (22)10;3) 220 = (25)4;2) 220 = (24)5;4) 220 = (210)2.176. 1) 0,01 = (0,1) ;2)223) 1925 ⎛ 5 ⎞==⎜ ⎟ ;16 16 ⎝ 4 ⎠( )= (c )177. 1) a 4 = a 23) c1025 2;4) x 203) (1,3 ⋅ 8) = 1,35 ⋅ 85 ;5179.

1) (ax ) = a 7 ⋅ x 7 ;73) (2,5cd )2 = 2,5 2 ⋅ c 2 ⋅ d 2 ;( )2= x2 ⋅ y6 ;2( )= (x )2) b6 = b3424) 0,0004 = (0,02);178. 1) (3 ⋅ 5) = 34 ⋅ 54 ;180. 1) xy 325 ⎛ 5 ⎞=⎜ ⎟36 ⎝ 6 ⎠210 22) (7 ⋅ 6) = 75 ⋅ 665⎛⎝1⎞7⎠3⎛ 1⎞⎝ 7⎠4) ⎜ 4 ⋅ ⎟ = 43 ⋅ ⎜ ⎟32) (6 y ) = 66 ⋅ y 664) (3mn) = 33 ⋅ m3 ⋅ n33( )32) a 2b = a 6 ⋅ b337( )3) 2b45(= 25 ⋅ b20 ;4) 0,1c3)2= (0,1) ⋅ c62() = 10 ⋅ n ⋅ m ; 2) (8a b ) = 8 ⋅ a3) ( −2 ,3a b ) = ( −2 ,3) ⋅ a ⋅ b ; 4) ( −2nm ) = ( −2)4181.

1) 10n2m3484 7 3123 4 2263123 484⋅ b21⋅ n4 ⋅ m12182. Если сторону квадрата увеличить в 2 раза, то S увеличится в 4раза; если в 3 раза, то S увеличится в 9 раз; если в 10 раз, то Sувеличится в 100 раз.183. Если ребро куба уменьшить в 2 раза, то V уменьшится в 8 раз;если в 10 раз, то V уменьшится в 1000 раз.184. 1) 45 ⋅ x5 = (4 x ) ;52) 23 ⋅ a 3 = (2a ) ;3) 5 4 ⋅ 7 4 = (5 ⋅ 7 )4;5) 16a 2 = (4a ) ;6) 81k 2 = (9 k ) ;34) 25 ⋅ 35 = (2 ⋅ 3) ;527) 97n7m7 = (9nm)7;8) 153a3b3 = (15ab)3( )= ( 5a ) ;( )2185. 1) c2 ⋅ d 10 = cd 5 ;3) 25a 42) a 4 ⋅ b6 = a 2b32 24) 81m2 = (9m)( )= (7x y )4 3 222(2186.