shestakov-all-gdz-2004 (546287), страница 7
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f(4)=16–81+5·4=–45;⎜⎝ 3 ⎠⎟x3x⎛ 2 ⎞2б) Обозначим ⎜ ⎟ = t , тогда: f(x)= 5 2 (t 3 − 1) :⎝5⎠3xxx2 ⋅ 5 x (t 2 + 1 + t )x5 2 (( t + 1) 2 + ( t − 1)2 )x5 2 ⋅ (t − 1)(t 2 + t + 1) ⋅ 5 2 ⋅ 2 ⋅ (t + 1)=− 4 ⋅ 2 2 = 5 x (t 2 − 1) − 4 ⋅ 2 2 =2 ⋅ 5x ⋅ (t 2 + t + 1)⎛ ⎛ 2 ⎞x⎞⎝⎠xx= 5 x ⋅ ⎜ ⎜ ⎟ − 1⎟ − 4 ⋅ 2 2 = 2 x − 5x − 4 ⋅ 2 2. f(2)=4–25–4·2=–29.⎜⎝ 5 ⎠⎟xxx⎛ x⎞⎛ x⎞ x x⎛ x⎞x⎜ 3 2 − 5 2 ⎟⎜ 3 2 + 5 2 ⎟ 5 4 ⋅ 3 4 ⎜ 3 4 + 5 4 ⎟⎜⎟⎜⎟⎜⎟34⎝⎠⎝⎠⎝⎠1.5.D06. а) f(x)=⋅⋅=xxxxxx 2⎛ x⎞⎛ x⎞22243 +53 ⎜3 + 5 ⎟5 4 ⋅ ⎜ 34 − 5 4 ⎟⎜⎟⎜⎟⎝⎠⎝⎠2xxxxx⎛ x⎞⎛ x⎞ x x⎛ x⎞ ⎛ x⎞⎛ x⎞⎜ 3 4 − 5 4 ⎟⎜ 3 4 + 5 4 ⎟ ⋅ 5 4 ⋅ 3 2 ⎜ 3 4 + 5 4 ⎟ ⋅ ⎜ 3 2 + 5 2 ⎟⎜ 34 + 5 4 ⎟⎜⎟⎜⎟⎜⎟ ⎜⎟⎜⎟⎠⎝⎠⎝⎠ ⎝⎠=⎝⎠=⎝;2xxxxxxxx⎛ x⎞⎛⎞⎛ x⎞⎛ x⎞⎛ x⎞4442442244⎜ 3 + 5 ⎟⎜ 3 − 5 ⎟3 ⋅ 5 ⎜ 3 + 5 ⎟⎜ 3 + 5 ⎟⎜ 3 − 5 ⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠f(8)=(32 + 52 )21156289;==−26344(3 + 52 )(32 − 52 ) −6586 ⋅168xxx⎛ x⎞⎛ x⎞ x x⎛ x⎞x⎜ 9 2 − 2 2 ⎟⎜ 9 2 + 2 2 ⎟ 2 4 ⋅ 9 4 ⎜ 2 4 − 9 4 ⎟⎜⎟⎜⎟⎜⎟94⎝⎠⎝⎠⎝⎠б) f(x)=⋅⋅=xxxxxx 2⎛ x⎞⎛ x⎞2 + 222494449 ⎜9 − 2 ⎟2 ⋅⎜9 + 2 ⎟⎜⎟⎜⎟⎝⎠⎝⎠2xxxxx⎛ x⎞⎛ x⎞⎛ x⎞ x x⎛ x⎞⎛ x⎞− ⎜ 94 − 24 ⎟⎜ 9 4 − 2 4 ⎟⎜ 9 4 + 2 4 ⎟⎜ 9 2 + 2 2 ⎟ ⋅ 2 4 ⋅ 9 2 ⎜ 2 4 − 9 4 ⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎠⎝⎠⎝⎠⎝⎠=⎝⎠=⎝;xxxxxxx 2⎛ x⎞⎛ x⎞⎛ x⎞⎛ x⎞⎛ x⎞4 ⎟⎜ 9 4 + 2 4 ⎟244224492−⎜9 ⋅ 2 ⎜ 9 − 2 ⎟⎜ 9 + 2 ⎟⎜ 9 + 2 ⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠54x− 4 ⋅ 22 =−(9 − 2) 24977=−.=−=−11 ⋅ 655911⋅ 93710307(9 − 2)(9 + 2)f(4)=41.5.D07.
а) f3(1)–f3(2)+f3(3)+…+(–1)n–1f3(n)+…==(0,1)3–(0,1)6+(0,1)9+…+(–1)n–1(0,1)3n+…=(0,1)30,0011==;1 − (−0,1)3 1, 001 1001б) f2(1)+f2(2)+f2(3)+…+f2(n)+…=(0,2)2+(0,2)4+(0,2)6+…+(0,2)2n+…==(0, 2)20,04 1== .1 − (+0, 2)2 0,96 241.5.D08. а) f2(–1)+f2(–2)+f2(–3)+…+f2(–n)+…=2⎛1⎞1⎜ ⎟14⎠⎛1⎞ ⎛1⎞⎛1⎞⎝16== ;= ⎜ ⎟ + ⎜ ⎟ + ... + ⎜ ⎟ + ... =21515⎝4⎠ ⎝4⎠⎝4⎠⎛1⎞1− ⎜ ⎟16⎝4⎠22n4б) f3(–1)–f3(–2)+f3(–3)+…+(–1)n–1f3(–n)+…=3⎛1⎞3⎛1⎞6⎛1⎞9⎛1⎞= ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ + ... + (−1)n−1 ⎜ ⎟⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠⎝ 3⎠3n⎛1⎞1⎜ ⎟13= ⎝ ⎠ 3 = 27 = .28281⎛ ⎞1− ⎜ − ⎟27⎝ 3⎠1.5.D09.а) 52x+52y+25x·5y–25y·5x=(5x–5y)2+2·5x+y+5x+y(5x–5y)=9+2·125+125·3=634;б) 32x+32y+9x·3y+9y·3x=32x+32y+32x+y+3x+2y=32x+32y+32+x+32+y==32x+32y+9(3x+3y)=(3x+3y)2–2·3x+y+90=102–2·32+90=190–18=172.⎛2⎞x1.5.D10.
а) Обозначим ⎜ ⎟ = t , тогда⎝7⎠⎛⎝f(x)= ⎜ 5t 2 + 11t − 8 +5 ⎞t + 11 ⎞ ⎛⎟ : ⎜ 5t − 4 +⎟=t +1 ⎠ ⎝t +1 ⎠⎛ 5t 3 + 16t 2 + 4t + 3 ⎞ ⎛ t + 1 ⎞ (t + 3)(5t 2 + t + 1)⎟⎟ ⋅ ⎜⎝ 5t 2 + t + 1 ⎟⎠ = (5t 2 + t + 1) = t + 3 =t +1⎝⎠= ⎜⎜⎛2⎞x⎛7⎞2491= ⎜ ⎟ + 3 ; f(–2)= ⎜ ⎟ + 3 = + 3 = 15 ;44⎝7⎠⎝2⎠⎛5⎞xt +4⎞ ⎛⎛31 ⎞б) ⎜ ⎟ = t , тогда f(x)= ⎜ 2t 2 + t + 2 +⎟ : ⎜ 2t − 7 +⎟=t +4⎠ ⎝t +4⎠⎝2⎠⎝x=(2t 2 + t + 3)(t + 4)22⎛5⎞= t + 4 = ⎜ ⎟ + 4; f(–1)= + 4 = 4 .2255(2t + t + 3)⎝ ⎠1.5.D11.а)322811234210(3 )=(2 )2 113 10114114(9 )=(8 )1110114114> 1 , так что 322811> 234210;55б)3230623455(3 )=(2 )1152 61153 5(9 )=(8 )65115115> 1 , так что 323071.5.D12.
а) Надо сравнить573553355351⎛5⎞= ⎜ ⎟ ⋅ 7−⎝ 3⎠ 35.35с 1. Возведем в квадрат.1⎛7⎞ 1< ⎜ ⎟ ⋅ 0,5 < 2 ⋅ = 1 . Значит,552⎝ ⎠37> 234521⎛7⎞= ⎜ ⎟ ⋅ 5−5⎝ ⎠ 5б) Надо сравнить5673< 55;5с 1. Возведем в квадрат.7575> ⋅1 > 1 . Значит, 5 < 3 .3§ 6. Логарифмические выраженияУровень А.(1.6.А01.
а) 8log8535) = ( 5)333(= 5; б) 6log653) = ( 3)555= 3.1.6.А02. а) log636+log232=2+5=7; б) log525+ log327=2+3=5.1.6.А03. а) log2781+ log279=424 2log33+ log33= + =2;333 3б) log168+ log1632= log16(8·32)= log16256=2.1.6.А04. а) log354– log32= log35424= log327=3; б) log224– log26= log2=26= log24=2.(1.6.А05. а) 5log3 5(б) 4log7 3)log 4 7()log5 3= 4log 4 7log3 5⋅= 5log3 5⋅log5 3 = 5)log 7 31log3 5= 51 = 5;= (7)log7 3 = 3.1.6.А06.а) log999+ log9911= log99(9·11)=1; б) log123+ log124= log12(3·4)= log1212=1.Уровень В1.6.В01.а)log16 4 + log16 24 − log16 6 = log164 ⋅ 24= log16 16 = 1;632 ⋅14= log 4 64 = 3.71111.6.В02. а) log16log381= log164= log44= ; б) log27log464= log273= .2231111.6.В03.
а) log4+ log5+ log3 =–3–2–2=–7;64259111б) log3 – log4 + log2 =–4+2–3=–5.81168б)56log 4 32 + log 4 14 − log 4 7 = log 41.6.В04. а) log18126– log187= log18126= log1818=1;7120= log1515=1.8б) log15120– log158= log1511.6.В05. а) log2004tg45º+ log 1 cos45º= log20041+ log 1241.6.В06.
а)7221 1= ;2 231 1=0+ = .2 22б) log2003ctg45º+ log 3 cos30º= log20031+ log 3log 7 5=0+4+ 6log6 2 + 2log2 18 = 5 + 2 + 18 = 5;б) 9log9 2 + 8log8 6 + 5log5 41 = 2 + 6 + 41 = 7.1.6.В07.а) log381+ log416+ log636=4+2+2=8; б) log327– log264+ log525=3–6+2=–1.135315751.6.В08. а) log3 3 (9 3) = log 3 ( 3)5 = ; б) log 4 2 (8 2) = log 2 ( 2)7 = .1.6.В09. а) 92 − log3 5 = 34 − 2 log3 5 = 34 − log3 25 =б) 43− log2 3 = 26 − 2 log2 3 =262⎛1⎞⎝ ⎠1.6.В10. а) 64log8 2 − ⎜ ⎟5⎛1⎞⎝ ⎠б) 64log8 3 − ⎜ ⎟51.6.В11. а)б)log5 8=3=816=3 ;2525641=7 .99log5= 82 log8 2 − 5− log5 8 = 8log8 4 − 5log5= 82 log8 3 − 5− log5 8 = 8log8 9 − 518= 9−18= 4−17=3 ;8817=8 .882log5 2522223log 2 163434= log 2 == 2; б) log 9 = log 3 == 33 = 27.log81 4949723997725log5 6 − 4log 4 32 = 52 log5 6 − 4log 4 32 = 36 − 32 = 2;1.6.В12.
а)4log5 8log 2 934log3 2549log7 5 − 3log3 9 = 4 49log 49 25 − 3log3 9 = 4 25 − 9 = 4 16 = 2.Уровень С.1.6.С01. а)б)( 19 )( 11 )log19 49+ 101.6.С02. а) log= logsin2π5= 1 + loglog11 25sin+6log 10 112π5=log 6 11=( 19 )( 11)log 19 7log 11 5+ 6log6 121 = 5 + 121 = 126;+ 10log10 121 = 7 + 121 = 128.π⎞π⎛ 3⎜ 2 3 cos ⎟ + logsin 2 π sin =5⎠5⎝5ππ3 ⎞2π⎛⎛⎞3 ⎟ = log 2 π ⎜ sin ⋅ 3 3 ⎟ =⎜ 2cos sinsin555⎝⎠⎠5 ⎝sin2π531113 = 1 + log 2 π 3 = 1 += 1+ ;2π3 sin 53b3log3 sin5572⎛π⎞π2π 3 ⎞⎛ 3⋅7 ⎟ =⎜ 2 49 cos ⎟ + log sin 2 π sin = log sin 2 π ⎜⎜ sin⎟3⎠1313⎝1313 ⎝⎠222= 1 + log 2 π 7 = 1 += 1+ .2π3 sin 133b3log 7 sin13б) log2πsin1311.6.С03. а) log 1 log 25 5 − 9 log5 3 = log 122б) log 1 log 27 3 − 161log 5 41= log 1 − 42 log 4 5 = 1 − 25 = −24.3331.6.С04. а) 32log0,5б) 64log 0,25 3 47=45451 2 log3 5−3= 1 − 25 = −24;2= 323⋅ log 1 3 47log 1 45=44321;45= 32− log32 5 =log 1 47=441.6.С05.
а) –log3log9 27 3 9 = – log3log 4147=1.471=4;811=log464=3.64log3 12 + log 4 12111.6.С06. а)=+= log12 4 + log12 3 = log12 12 = 1;log 3 12 ⋅ log 4 12 log 4 12 log3 12log 2 18 + log 9 1811б)=+= log18 9 + log18 2 = log18 18 = 1.log 2 18 ⋅ log 9 18 log 9 18 log 2 18б) –log4log8 16 4 8 =–log41.6.С07. а) 9б) 4log3 5 + 2 log 1 4log 2 5 + 4 log 1 3169=4log 2 5 + log 1 321.6.С08. а) 25log16 2 + log5б) 49log81 3+ log716= 32 log3 5 ⋅ 9=417= 25 4log 22 log953=4+ log 25 71+ log 49 614= 25 ⋅log 425919=1 ;1616=257=2 .99= 4 25 ⋅ 7 = 7 5;+ log 49 61= 49 4= 49 4= 49 4 ⋅ 49log49 6 = 7 ⋅ 6 = 6 7.log 3 6 log 3 181.6.С09. а)−= log 32 6 − log 3 18 ⋅ log 3 2 =log 6 3 log 2 3=(1+log32)2–(2+log32)log32=1;б)log3 63 log 3 21 log 3 63−=− log 32 21 =log 7 3 log 21 3log 7 3= log3 7 ⋅ ( 2 + log 3 7 ) − (1 + log3 7 ) = 2 log 3 7 + log 32 7 − 1 − 2 log 3 7 − log32 7 =–1.21616161.6.С10.
а) log a 6 ab = log a ab = (1 + log a b) = (1 + 29) = 5;б) log a 358a 1a 11= log a = (1 − log a b) = (1 + 11) = 4.3b 3b 31.6.С11. а) 2 log 1211− 3log8 35 = log 1− log 2 35 = log 2 36 − log 2 35 > 0 , так что636212 log 1 > 3log8 35 ;62б)2 log 1211− 5log 32 26 = log 1− log 2 26 = log 2 25 − log 2 26 < 0 ,5252такчто1< 5log 32 26 .522 log 111.6.С12.
а) 49 31б) 36 4log 7 27 + 2 log 7 6log6 16 + 4 log6 4 2= 49log7 3+ log7 6 = 49log7 18 = 49log49 324 = 324;= 36log6 2 + log6 2 = 36log6 4 = 36log36 16 = 16.Уровень D.1.6.D01. а) 2 log 2325+ 6+ log 2 (11 + 2 30) = 2 log 2325+ 6+ log 2 ( 5 + 6)2 == 2 log 2 32 − 2 log 2 ( 5 + 6) + 2 log 2 ( 5 + 6) = 10;б) 2 log397+ 6+ log 3 (13 + 2 42) = 2log 3 9 − 2 log 3 ( 7 + 6) ++ log3 ( 7 + 6)2 = 2 ⋅ 2 − 2 log3 ( 7 + 6) + 2 log3 ( 7 + 6) = 4.26 log 7 2 +log 7 320log 7 26 + log 7 5log5 71.6.D02.
а) log70320====log 7 70 log 7 7 + log 7 2 + log 7 5 1 + log 2 + 17log 5 71a = 6ab + 1 ;=1 a + ab + 11+ b +a6b +26 log 5 2 +log 5 576log 5 26 + log 5 32log3 5б) log30576====log 5 30 log 5 5 + log 5 2 + log5 3 1 + log 2 + 15log 3 526ab + 2a==.1 a + ab + 11+ b +alog 3 153 log 3 459−1.6.D03. а)=(log39+ log317)( log33+ log317)–( log317+log 51 3log17 36b ++log327)·log317= 2 log32 3 + 3log3 17 + log 32 17 − log32 17 − 3log3 17 = 2;б)log 2 176 log 2 352=(log216+log211)(log22+log211)–−log 22 2log11 259–(log211+log232)log211= 4 log 22 2 + 5log 2 11 + log 22 11 − log 22 11 − 5log 2 11 = 4.(1.6.D04.
а) 32 + log3 5 + 4(б) 32 + log3 5 − 9)log 6 7)log 7 9= (9 ⋅ 5 + 4)log49 81 = 49log49 81 = 81;= (9 ⋅ 5 − 9)log36 49 = 36log36 49 = 49.()1.6.D05. а) 21 − 22 + log2 5 log5 3 3 ⋅ log3 125 = (21 − 4 ⋅ 5) log5 3 3 ⋅ log3 125 =⎛1⎞= ⎜ log5 3 ⎟ ⋅ ( 3log3 5 ) = 1;⎝3⎠()14б) 22 − 51+ log5 4 log 2 4 3 ⋅ log3 16 = (22 − 5 ⋅ 4) ⋅ log 2 3 ⋅ 4 log3 2 = 2.log13 3log11 13log3 1113log11 3 ⋅11log3 13 ⋅ 3log13 11 13 log13 11 ⋅11 log11 3 ⋅ 3 log3 131.6.D06. а) log 11 log 3 log 13 = log 11 log 3 log 13 =13 3 ⋅11 13 ⋅ 3 1113 3 ⋅11 13 ⋅ 3 111=113log13 11 ⋅13 log11 3 ⋅11log3 13131log11 31log3 13⋅11⋅31log13 11= 1;log19 5log17 19log5 1719log17 5 ⋅17log5 19 ⋅ 5log19 17 19 log19 17 ⋅17 log17 5 ⋅ 5 log5 19б) log 17 log 5 log 19 = log 17 log 5 log 19 =19 5 ⋅17 19 ⋅ 5 1719 5 ⋅17 19 ⋅ 5 171=115 log19 17 ⋅19 log17 5 ⋅17 log5 19191log17 5⋅171log5 19⋅51log19 17= 1.1.6.D07.
а) 6lg(4–2 3 )–12lg( 3 –1)=6lg(4–2 3 )–6lg(4–2 3 )=0;б) 5lg(4+2 3 )–10lg( 3 +1)=5lg(4+2 3 )–5lg( 3 +1)2=5lg(4+2 3 )––5lg(4+2 3 )=0.3511⋅ log 5 = log 3 log 5 = log35·log53=1;1515534911б) (1–log436)(1–log936)= log 4 ⋅ log9 = log 4 log9 = log49·log94=1.363694log 2 14 log 2 7−= (log22+log27)(log24+log27)–(log27+log28)log27=1.6.D09.
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