shestakov-all-gdz-2004 (546287), страница 6
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Степеньс действительным показателемУровень А.1.5.А01.⎛⎛ 1 ⎞⎞2⎛⎛ 1 ⎞⎞4⎛1⎞21⎞⎛⎜ −2 + ⎟ ⋅ 22⎠а) ⎜ f (−2) f ⎜ ⎟ ⎟ = ⎜ (6)−2 ⋅ (6) 2 ⎟ = 6⎝⎟⎝ 2 ⎠ ⎠ ⎝⎜⎝⎠⎛1⎞41⎞⎛⎜ −1+ ⎟ 44⎠б) ⎜ f (−1) f ⎜ ⎟ ⎟ = ⎜ 7 −1 ⋅ 7 4 ⎟ = 7⎝⎟⎝ 4 ⎠ ⎠ ⎜⎝⎝⎠1.5.А02.= 6−3 == 7 −3 =21;2161.3432⎛ 7 ⋅ 22 х + 5 ⋅ 2−2 х ⎞ ⎛ 7 ⋅ 22 х − 5 ⋅ 2−2 х ⎞⎟⎟ − ⎜⎜⎟⎟ =22⎝⎠ ⎝⎠а) f2(x)–g2(x)= ⎜⎜=49 ⋅ 24 х + 70 + 25 ⋅ 2−4 х 49 ⋅ 24 х − 70 + 25 ⋅ 2−4 х 140−== 35;44422⎛ 3 ⋅ 52 х − 4 ⋅ 5−2 х ⎞ ⎛ 3 ⋅ 52 х + 4 ⋅ 5−2 х ⎞⎟⎟ − ⎜⎜⎟⎟ =22⎝⎠ ⎝⎠б) f2(x)–g2(x)= ⎜⎜=9 ⋅ 54 х − 24 + 16 ⋅ 5−4 х 9 ⋅ 54 х + 24 + 16 ⋅ 5−4 х −48−== −12.4441.5.А03.а) 5f(3)+9f(2)+7f(1)+2f(0)=5·(0,1)3+9·(0,1)2+7·0,1+2==0,005+0,09+0,7+2=2,795;б) 6f(3)+9f(2)+4f(1)+4f(0)=6·(0,1)3+9·(0,1)2+4·(0,1)+4==0,006+0,09+0,4+4=4,496.1.5.А04.а) 5f(–3)+8f(–2)+f(–1)+2f(0)=5·10–3+8·10–2+10–1+2==0,005+0,08+0,1+2=2,185;б) 5f(–3)+2f(–2)+2f(–1)+4f(0)=5·10–3+2·10–2+2·10–1+4==0,005+0,02+0,2+4=4,225.1.5.А05.⎛1⎞⎝ ⎠⎛1⎞−2а) f −2 ⎜ ⎟ + g 5 (−0, 2) = ⎜ 6 2 ⎟ + ( (0,1) −0,2 ) = 6−1 + 0,1−1 = + 10 = 10 ;⎜ ⎟266⎛1⎞⎝ ⎠⎝151⎠⎛1⎞−3б) f −3 ⎜ ⎟ + 2 g 4 (−0, 25) = ⎜ 33 ⎟ + 2 ⋅ ( (0, 2) −0,25 ) = 3−1 + 2 ⋅ 0, 2−1 =⎜ ⎟3⎝4⎠11= + 10 = 10 .331.5.А06.а) f(x)=5x·0,22x=5x·0,04x=(0,2)x, основание функции — 0,2;б) f(x)=102x·0,13x=100x·(0,001)x=(0,1)x, основание функции —0,1.
Уровень В.47−1.5.В01. а) f(x)=72x· 81⎛1⎞⎝ ⎠х2⎛1⎞х⎛ 49 ⎞х49=49x· ⎜ ⎟ = ⎜ ⎟ , основание функции —;9⎝9⎠ ⎝ 9 ⎠49 7= ;93f⎜ ⎟=2б) f(x)=43x· 64−х2⎛1⎞⎝ ⎠( )1.5.В02. а) f(x)= 3 32х()2х⎛1⎞⎝ ⎠х⎛1⎞⋅ 9−0,5 х = 27 х ⋅ ⎜ ⎟ = 9 х , основание функции — 9;⎝ 3⎠⎛1⎞f ⎜ ⎟ = 9 = 3;⎝ 2⎠б) f(x)= 4 2х=64x· ⎜ ⎟ = 8 х , основание функции — 8; f ⎜ ⎟ = 3 8 = 2.38х⎛1⎞⋅16−0,25 х = (32) х ⋅ ⎜ ⎟ = 16 х , основание функции — 16;⎝2⎠⎛1⎞f ⎜ ⎟ = 4 16 = 2.⎝4⎠1.5.В03. а) f(x)=3x +1 + 3x + 2 3x +1 (1 + 3) ⎛ 4 ⎞ ⎛ 3 ⎞== ⎜ ⎟⋅⎜ ⎟4 x + 2 − 4 x +1 4 x +1 (4 − 1) ⎝ 3 ⎠ ⎝ 4 ⎠основание функции —3⎛3⎞; 9f(–1)= 9 ⋅ ⎜ ⎟4⎝4⎠−1= 9⋅x +1x⎛ 3⎞=⎜ ⎟ ,⎝ 4⎠4= 12;3x4 x +1 + 4 x + 2 4 x (4 + 16) ⎛ 4 ⎞4== ⎜ ⎟ , основание функции — ;55 x + 2 − 5 x +1 5 x (25 − 5) ⎝ 5 ⎠516f(–1)= 16 ⋅ = 20.4б) f(x)=x3x +1 + 3х + 3 + 3x + 2 3x (3 + 27 + 9) ⎛ 3 ⎞= x=⎜ ⎟ ,5x + 2 + 14 ⋅ 5x5 (25 + 14) ⎝ 5 ⎠325основание функции — ; 9f(–2)= 9 ⋅ = 25;591.5.В04.
а) f(x)=x4 x +1 + 4 x + 2 + 4 х + 3 4 x (4 + 16 + 64) ⎛ 4 ⎞4== ⎜ ⎟ , основание функции —;77 x + 2 + 35 ⋅ 7 x7 x (49 + 35)⎝7⎠49491= 12 .4f(–2)= ⋅ 4 =1644б) f(x)=21.5.В05. а) f(2x)–8g2(x)=–⎛ 5 х − 5− х ⎞52 х + 5−2 х52 х + 5−2 х− 8 ⋅ ⎜⎜−⎟ =⎟88 ⎠8⎝52 х − 2 + 5−2 х 2 1= = ;88 42б) f(2x)–14g2(x)=48⎛ 2 х − 2− х ⎞22 х + 2−2 х22 х + 2−2 х− 14 ⋅ ⎜⎜−⎟ =⎟1414⎝ 14 ⎠–22 х − 2 + 2−2 х2 1== .1414 721.5.В06. а) g(2x)–6g2(x)=–⎛ 4 х + 4− х ⎞42 х + 4−2 х42 х + 4−2 х− 6 ⋅ ⎜⎜−⎟⎟ =666⎝⎠42 х + 2 + 4−2 х21=− =− ;6632⎛ 7 х + 7− х ⎞7 2 х + 7 −2 х7 2 х + 7 −2 х 7 2 х + 2 + 7 −2 х2− 2 ⎜⎜−= − = −1.⎟⎟ =22222⎝⎠1а 2аа1.5.В07.
а) 6 − а = 6; (6 ) –6·(6 )–1=0;6б) g(2x)–2g2(x)=6а=3+ 10 (так как 6а>0). Тогда (6а–6)6а=( 10 –3)( 10 +3)=10–9=1;б) 4а +1= 4; (4а)2–4·(4а)+1=0; 4а=2± 34а(4а–4)4а=( 3 –2)( 3 +2)=3–4=–1 или(4а–4)4а=(–2– 3 )(2– 3 )=(2+ 3 )( 3 –2)=3–4=–1. Так что (4а–4)4а=–1.11а+2b=–4 и f(1)=–2, то есть 5а+ b=–2, так что5220⎧a=−⎧а + 10b = −20⎧99b = −196⎪⎪99;;;⎨⎨⎨⎩10a + b = −4⎩99a = −20⎪b = −1 97⎪⎩9911б) f(–1)=1, то есть а+5b=1; f(1)=–4, то есть 3а+ b=–4, так что3565⎧⎪⎪b = 224⎧а + 15b = 3⎧224b = 65; ⎨; ⎨.⎨⎩15a + b = −20⎩224a = −303⎪a = −1 79⎪⎩2241.5.В08. а) f(–1)=–4, то есть⎛ 232 (23 ) 2 ⎞⎟1.5.В09. а) f(2)= ⎜⎜ (232 ) 2 ⎟⎝⎠−2⎛ 29 ⋅ 2 6 ⎞= ⎜⎜ 18 ⎟⎟⎝ 2 ⎠−2= (2−3 ) −2 = 26 = 64;1⎛ 332 ⋅ (33 )2 ⎞ 39 ⋅ 36 3151⎟ = 18 = 18 = 3−3 = .⎜ (332 )2 ⎟2733⎝⎠б) f(–1)= ⎜⎛ 6−32 (63 )−2 ⎞⎟1.5.В10.
а) f(2)= ⎜⎜ (63−2 )18 ⎟⎝⎠⎛ 3−23 ⋅ (32 )−3 ⎞⎟б) f(6)= ⎜⎜ (32−3 )16 ⎟⎝⎠−616−217⎛ 6−9 ⋅ 6−6 ⎞= ⎜⎜2⎟⎟⎝ 6⎠⎛ 3−8 ⋅ 3−6 ⎞= ⎜⎜⎟⎟2⎝ 3⎠−616−217= (3−16 )−= (6−17 )616−217= 62 = 36;= 36 = 729.491.5.В11. а) f(44)=744=(74)11, g(33)=833=(83)11; h(22)=922=(92)11.Так что f(44)>g(33)>h(22);б) f(60)=560=(54)15; g(45)=745=(73)15; h(30)=330=(32)15, так чтоf(60)>g(45)>f(30).1.5.В12. а) f(160)=5160=2580<2780=3240=g(240). f(160)<g(240);б) f(270)=5270=(125)90<(1024)90=4450=g(450). f(270)<g(450).Уровень С.⎛1⎞⎛1⎞1.5.С01. а) f ⎜ ⎟ − g ⎜ ⎟ − g (3) = 11 − 5 3 − 27 < 0;⎝2⎠⎝5⎠⎛1⎞⎛1⎞б) f ⎜ ⎟ − g ⎜ ⎟ − g (2) = 5 17 − 3 4 − 16 < 0.⎝5⎠⎝ 3⎠141414⎛ 5 ⎞ 3 ⎛ 1−2 ⎞ 3 ⎛ 3 ⎞ 31.5.С02.
а) ⎜⎜ 7 ⎟⎟ = ⎜ 5 2 7 ⎟ = ⎜ 514 ⎟ = 5;⎜ ⎟⎜⎟⎝ 25 ⎠⎝ ⎠⎝⎠⎛ 34 ⎞б) ⎜⎜⎟⎟⎝ 64 ⎠−67⎛ 1−3 ⎞= ⎜ 43 2 ⎟⎜⎟⎝⎠−67⎛ −7 ⎞= ⎜4 6 ⎟⎜⎟⎝⎠−67= 4.1.5.С03. а) f2(17)+f2(–17)=(317)2+(3–17)2=(317+3–17)2–2=(f(17)+f(–17))2–2=a2–2;б) f2(24)+f2(–24)=(724)2+(7–24)2=(724–7–24)2+2=(f(24)–f(–24))2+2=a2+2.xy1.5.С04.
а) 4·6 +3·6 =4·xyб) 2·3 –5·3 =2·х+ у3 2х+ у6 2х− у⋅3 2−5х− у⋅6 23х+ у2х− у3 2+ 3⋅6х+ у2х− у6 2a= 4ab + 3 ;ba= 2ab − 5 .b2 a + 4 ⋅ 2b2a − b + 42b (2a − b + 4);=−7= −7;7;=−2 a − 2 ⋅ 2b2a − b − 22b (2 a − b − 2)22a–b+4=–7·2a–b+14; 8·2a–b=10; 2a–b=1 ;8abba −bб) 7 + 3 ⋅ 7 = 7 (7 + 3) = 2; 7a–b+3=2·7a–b+2; 7a–b=1.7 a + 7b7b (7 a − b + 1)1.5.С06. а) f(–1)+f(–2)+f(–3)+…+f(–n)+…= 1 + 1 + 1 + 1 + ... + 1 + ...
=6 62 63 646n1= 6 = 1;1 51−6б) f(–1)–f(–2)+f(–3)+…+(–1)n–1f(–n)+…= 1 − 1 + 1 + ... + (−1) n 1 + ... =5 52 535n115== .⎛ 1⎞ 61− ⎜ − ⎟⎝ 5⎠1.5.С05. а)501.5.С07. а) f(3)+f(6)+f(9)+…+f(3n)+…=0,43+0,46+0,49+…+0,43n+…==0, 430, 0648==;1 − 0, 43 0,936 117б) f(2)+f(4)+f(6)+…+f(2n)+…=0,32+0,34+0,36+…+0,32n+…==0,320, 099== .21−0,09911 − (0,3)1.5.С08.
а) f(1)+f(3)+f(5)+…+f(2n–1)+…=0,3+0,33+0,35+…+0,32n–1+…==0,30,3 30== ;1 − 0,32 0,91 91б) f(1)+f(4)+f(7)+…+f(3n–2)+…=0,2+0,24+0,27+…+0,23n–2+…==0, 20, 225.==1 − 0, 23 0,992 1241.5.С09. а) f(–1)–f(–3)+f(–5)+…+(–1)n–1f(–2n+1)+…=1 1 11= − 3 + 5 + ... + (−1)n −1 2 n −1 + ... =3 3 331133= 3 = ;⎛ 1 ⎞ 10 101− ⎜ − ⎟⎝ 9⎠ 9б) f(–3)–f(–7)+f(–11)+…+(–1)n–1f(–4n+1)+…=1 111= − 7 + 11 + ... + (−1) n −1 4 n −1 + ... =8 2 22333323n1812= 8 = .17117⎛⎞1− ⎜ − ⎟⎝ 16 ⎠ 161.5.С10. а) 5 2 ⋅ 5 2 ⋅ 5 2 ⋅ ... ⋅ 5 2 ⋅ ... = (5)б)34733323n⋅ 4 7 ⋅ 4 7 ⋅ ... ⋅ 4 7 ⋅ ... = (4)3 3 33+ + + ... + + ...2 22 232n3 3 33+ + + ... + + ...7 7 2 737n=3711−(4) 7=3211−(5) 2=37647=32152= 53 = 125;1= 4 2 = 2.(3x + 3− x )(3 y + 3− y ) (3 y − 3− y )(3x − 3− x )−=16162 ⋅ (3x − y + 3 y − x ) 1 3x − y + 3− ( x − y ) 115= ⋅= ⋅ f ( x − y) = ;=162422(4 x + 4− x )(4 y + 4− y ) (4 x − 4− x )(4 y − 4− y )−=б) f(x)f(y)–g(x)g(y)=36364 x + y + 4 x − y + 4 y − x + 4− x − y − 4 x + y + 4 x − y + 4 y − x − 4− x − y==361.5.С11.
а) f(x)f(y)–g(x)g(y)==4 x − y + 4 y − x 1 ⎛ 4 x − y + 4− ( x − y ) ⎞ 1= ⋅ ⎜⎜⎟⎟ = ⋅ f ( x − y ) = 3.183 ⎝6⎠ 322 y 2 + 51.5.С12. а) 311 y 22+=3511 y 2512122= 9 ⋅ 311 y = 9 ⋅ (35 )11 y = 9 ⋅ (11 35 ) y > 95115 x 2 −1125x2=28 y2 +3б) 34 y23−1115x2+=3⎛ 1 ⎞ x2= 8 ⋅ ⎜ 5 11 ⎟ < 8; Так что 3⎜ 2 ⎟⎝⎠234 y2= 9 ⋅ ( 27)48 y2 +3так что 34 y21.5.D01.
а)19 x2 − 4y23 x2>9,а 222 y 2 + 511 y 2=23−15 x 2 −11>243 x25 x2, для всех х и у;1⎛ 1 ⎞ x2= 8 ⋅ ⎜⎜ 3⎟⎟ < 8;⎝ 16 ⎠9 x2 − 4>23 x2, для всех х и у.Уровень D.f (61) 661 661 361 3 ⋅ 360 3 ⋅ (34 )15 3 ⋅ (81)15======>1 ,g (76) 476 2152 291 2 ⋅ 290 2 ⋅ (26 )15 2 ⋅ (64)15так что f(61)>g(76);б)f (33) 633 633 333 3 ⋅ 332 3 ⋅ 916=====> 1 , так что f(33)>g(41).g (41) 441 282 249 2 ⋅ 248 2 ⋅ 8161.5.D02. а) (5–52х)2·5–х+(5–5–2x)·5x=25·5–x–10·5x+53x+25·5x–10·5–x+5–3x==15(5x+5–x)+53x+5–3x=15·5+(5x+5–x)(52x–1+5–2x)==75+5·(1+(5x+5–x)2–2)=75+5·(25–3)=185;б) (4+22x)2·2–x+(4+2–2x)2·2x=16·2–x+8·2x+23x+16·2x+8·2–x+2–3x==24(2x+2–x)+23x+2–3x=24·4+(2x+2–x)(22x–1+2–2x)==96+4·((2x+2–x)2–3)=96+4(16–3)=148.2xx⎛⎛ ⎛ 3 ⎞2 x⎛3⎞⎛ 3⎞ ⎞⎞⎜25x ⎜ ⎜ ⎟ + 2 ⋅ ⎜ ⎟ + 1 − 8 ⋅ ⎜ ⎟ ⎟ ⎟2x⎜⎝ 5 ⎠⎞⎜⎛⎛ 3 ⎞⎝5⎠⎝ 5 ⎠ ⎟⎠ ⎟⎝1.5.D03.
а) f(x)= ⎜ ⎜ ⎜ ⎟ − 1⎟ −⎟ ⋅x⎜⎝ 5 ⎠⎟⎛ ⎛ 3⎞ ⎞⎜⎝⎟x⎠25 ⎜1 − ⎜ ⎟ ⎟⎜⎜⎟⎟⎜ ⎝5⎠ ⎟⎝⎠⎝⎠22xx 3xx⎞⎛⎛⎞⎜ ⎜1− ⎛ 3 ⎞ ⎟ − ⎛ 3 ⎞ − 2⎛ 3 ⎞ −1+ 8⎛ 3 ⎞ ⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜ ⎜ ⎝ 5⎠ ⎟ ⎝ 5⎠⎝ 5⎠⎝ 5⎠ ⎟54x⎠⎟ ⋅⋅=⎜⎝2x3x4xx⎛ ⎛ 3⎞⎛ ⎛ 3⎞ ⎞⎟⎛ 3⎞ ⎛ 3⎞ ⎞ ⎜4x5 ⋅⎜9⋅ ⎜ ⎟ − 6⋅ ⎜ ⎟ + ⎜ ⎟ ⎟ ⎜⎜1− ⎜ ⎟ ⎟⎟⎜ ⎝ 5⎠⎟ ⎜⎜ ⎝ 5⎠ ⎟55⎟⎝⎠⎝⎠⎝⎠ ⎝⎝⎠⎠2⋅1x⎛ 3⎞ ⎛ ⎛ 3⎞ ⎞3−⎜⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎟⎝5⎠ ⎝ ⎝5⎠ ⎠2x2x⎛ 3 x ⎛ 3 2x⎞⎞3⎜ − ⎜⎛ ⎟⎞ ⎜ ⎜⎛ ⎟⎞ − 2 ⎜⎛ ⎟⎞ − 3 ⎟ ⎟⎟⎟⎜ ⎝ 5 ⎠ ⎜⎝ ⎝ 5 ⎠⎝5⎠⎠ ==⎜xxx ⎟⎜ ⎛ 3 ⎞ ⎛ ⎛ 3 ⎞ ⎞⎛ ⎛ 3 ⎞ ⎞ ⎟3−⎟⎜⎜ ⎜ 5 ⎟ ⎜⎜1 − ⎜ 5 ⎟ ⎟⎜⎟⎜ ⎜ 5 ⎟ ⎟ ⎟⎟⎝ ⎝ ⎠ ⎝ ⎝ ⎠ ⎠⎝ ⎝ ⎠ ⎠ ⎠22⎛ ⎛ ⎛ 3 ⎞xx⎞⎛ 3 x ⎞ ⎞⎜ − ⎜ ⎜ ⎟ − 3 ⎟⎜ ⎛⎜ ⎞⎟ + 1⎟ ⎟ ⎛⎜ ⎛ 3 ⎞ + 1 ⎞⎟⎜⎟⎜⎟⎜⎟xx 2⎜ ⎝5⎠⎠⎝ ⎝ 5 ⎠⎠⎟ = ⎜⎝5⎠⎟ = ⎛3 +5 ⎞ .=⎜ ⎝⎜⎟x ⎟⎜ x x ⎟⎟xx⎜⎜ ⎛ ⎛ 3 ⎞ ⎞⎛ ⎛ 3 ⎞ ⎞ ⎟ ⎜ 1 − ⎛ 3 ⎞ ⎟ ⎝ 5 − 3 ⎠13−−⎜⎟⎜⎟⎜⎜ ⎜ ⎜⎝ 5 ⎟⎠ ⎟⎜ ⎜⎝ 5 ⎟⎠ ⎟ ⎟⎟ ⎜⎝ ⎜⎝ 5 ⎟⎠ ⎟⎠⎠⎝⎠ ⎠⎝ ⎝522222⎛ 311 + 511 ⎞ ⎛ 3−11 + 5−11 ⎞⎛ 311 + 511 ⎞ ⎛ 511 + 311 ⎞−=⎜⎟⎜⎜ 11 11 ⎟⎟ − ⎜⎜ 11 11 ⎟⎟ = 0;1111 ⎟1111−−⎟ ⎜⎟⎝ 5 −3 ⎠ ⎝ 5 −3 ⎠⎝ 5 −3 ⎠ ⎝ 3 −5 ⎠Так что f(11)–f(–11)= ⎜⎜22⎛x⎞⎞2x⎛⎛⎞⎛7⎞x⎜ ⎛7⎞⎜41−− 4⋅⎜ ⎟ ⎟ ⎟⎜⎟⎜⎟2⎜⎟⎜⎜ ⎝2⎠⎝2⎠ ⎟⎟⎛ ⎛ 7 ⎞x ⎞⎝⎠⎠⎟ ⋅б) f(x)= ⎜⎜ ⎜ ⎜ ⎟ − 1⎟ − ⎝x⎟⎜⎝ 2 ⎠⎟⎛⎞⎛7⎞⎠⎜⎝⎟4 x ⎜1 − ⎜ ⎟ ⎟⎜ ⎝2⎠ ⎟⎜⎟⎝⎠⎜⎟⎝⎠22xx 3xx⎞⎛⎛⎞⎜ ⎜1− ⎛ 7 ⎞ ⎟ − ⎛ 7 ⎞ + 2⎛ 7 ⎞ −1+ 4 ⋅ ⎛ 7 ⎞ ⎟⎜ ⎟ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎟4x⎜⎜⎝ 2⎠ ⎠ ⎝ 2⎠⎝ 2⎠⎝ 2⎠2⎟ =⋅=⎜⎝xxxx2х2x⎛⎞⎛⎞⎜⎟⎛7⎞ ⎛ 7⎞⎛ 7⎞⎛ 7 ⎞ ⎛ ⎛ 7 ⎞⎞4x ⎛ 7 ⎞2 ⋅ ⎜ ⎟ ⎜ 9 − 6⋅ ⎜ ⎟ + ⎜ ⎟ ⎟ ⎜⎟⎜ ⎟ ⎜ 3 − ⎜ ⎟ ⎟⎜1− ⎜ ⎟⎟⎟⎝ 2 ⎠ ⎝⎜⎝ 2 ⎠ ⎝ 2 ⎠ ⎠⎟ ⎜⎝ 2 ⎠ ⎝⎜ ⎝ 2 ⎠ ⎠⎟ ⎝ ⎝ 2 ⎠ ⎠⎝⎠2x2⎛ ⎛ 7 ⎞ x ⎛ ⎛ 7 ⎞2 x⎞⎞7⎛ ⎛ 7 ⎞x ⎞⎜ − ⎜ ⎟ ⎜ ⎜ ⎟ − 2 ⎛⎜ ⎞⎟ − 3 ⎟ ⎟+1⎜⎟xx 2⎜⎟⎜⎟⎜ ⎝ 2 ⎠ ⎝⎝ 2 ⎠⎝2⎠⎠⎟ = ⎜⎝2⎠⎟ = ⎛2 +7 ⎞ .=⎜⎜⎟xxxxxx⎜ ⎛ 7 ⎞ ⎟ ⎜ 2 − 7 ⎟⎟⎝⎠⎜ ⎛ 7 ⎞ ⎛ ⎛ 7 ⎞ ⎞⎛ ⎛ 7 ⎞ ⎞ ⎟⎜31−−⎜⎟⎜⎟⎜ 1 − ⎜ 2 ⎟ ⎟⎟⎜⎜ ⎜⎝ 2 ⎟⎠ ⎜ ⎜⎝ 2 ⎟⎠ ⎟⎜ ⎜⎝ 2 ⎟⎠ ⎟ ⎟⎟⎝⎠⎝⎠⎝⎠⎝⎠⎠⎝2222⎛ 2−4 + 7 −4 ⎞ ⎛ 24 + 7 4 ⎞⎛ 24 + 7 4 ⎞ ⎛ 24 + 7 4 ⎞⎟ − ⎜⎜ 4 4 ⎟⎟ = ⎜⎜ 4 4 ⎟⎟ − ⎜⎜ 4 4 ⎟⎟ = 0.−4−4 ⎟⎝ 2 −7 ⎠ ⎝ 2 −7 ⎠⎝ 7 −2 ⎠ ⎝ 2 −7 ⎠Так что f(–4)–f(4)= ⎜⎜хt −9⎛ 5⎞1.5.D04.
а) Обозначим ⎜ ⎟ = t , тогда f(x)=⋅14t −3 t +9⎝ ⎠=( t − 3)( t + 3) ⋅ ( t + 3)(t − 3 t + 9)(t − 3 t + 9)( t − 3)⎛ 4⎞⎛ 7⎞( t)3+ 27t −3−6( t)−9 =−1x⎛ 5⎞− 6 t − 9 = ( t + 3) 2 − 6 t − 9 = t = ⎜ ⎟ .⎝ 14 ⎠475Так что f ⎜ − ⎟ > f ⎜ − ⎟ , так как − < − , а<1;5914⎝ 5⎠⎝ 9⎠⎛2⎞х=( t − 3)( t + 3) ⋅ ( t − 3)(t + 3 t + 9)(t + 3 t + 9)( t + 3)⎛7⎞⎛ 2⎞( t)3t −9⋅б) Обозначим ⎜ ⎟ = t , тогда f(x)=t −3 t +9⎝7⎠+ 27t +3+6( t)−1−9 =x⎛2⎞+ 6 t − 9 = ( t − 3)2 + 6 t − 9 = t = ⎜ ⎟ .⎝7⎠722Так что f ⎜ − ⎟ > f ⎜ − ⎟ , так как − < − , а <1.1277⎝ 12 ⎠⎝ 7⎠x⎛ 2 ⎞21.5.D05.
а) Обозначим ⎜ ⎟ = t , тогда:⎝3⎠533xf(x)= 3 2 (t 3 − 1) :3xx2 ⋅ 3x (t 2 + t + 1)x3 2 ((+ 5 ⋅ 22 =t − 1) 2 + ( t + 1)2 )xxx3 2 ⋅ 3 2 ⋅ (t − 1)(t 2 + t + 1)(2t + 2)=+ 5 ⋅ 2 2 = 3x (t 2 − 1) + 5 ⋅ 2 2 =3x ⋅ 2 ⋅ (t 2 + t + 1)⎛ ⎛ 2 ⎞x⎞⎝⎠xx= 3x ⋅ ⎜ ⎜ ⎟ − 1⎟ + 5 ⋅ 2 2 = 2 x − 3x + 5 ⋅ 2 2.
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