| Atbilžu arhīvs | № 26041, Algebra, 10 klase uzdevums ir faila | | |
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angel | 3²=x+y lg(xy)=lg2³
9=x+y xy=8
x=9-y
y(9-y)=8 y²-9y+8=0 D=81-32=49 y=(9+7)/2=8 x=9-y=9-8=1 y=(9-7)/2=1 x=9-y=9-1=8 | | |
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vespertilio | 2y-x=1 x=4+y
2y-4-y=1 y=5 x=9 | | |
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BlackRock | {log3 (x+y)=2 {lg xv + lg y=3lg 2
log3 (x+y)=2 log3 (x+y)= log3 9 x+y=9
lg x + lg y=3lg 2 lg x + lg y=lg 23 lg x*y=lg 8 x*y=8
{x+y=9 {x*y=8
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| | № 26101, Algebra, 10 klase napiwite polnostju vse rewenie , tam tri primera | | |
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snow | 1) 2(x+5) > x -4 2x+10>x -4 2x - X > -4-10 X > -14 X pridnalizhit (-14;+∞) 2) D.A . 3x + 4 ≠0 x≠-4/3≠-1 1/3 odin i odna trec 2x²+x-6<=0 D = 1² -4 ·2 ·-6 = 1 + 48 = 49 X1 = (-1-7)/4 = -8/4 = -2 X2 = (-1+7)/4 = 6/4 = 1,5 3) D.A. 1-x≠0 x≠1 x+4≠0 x≠-4 (x+4)/(-x²-3x+4) + (2-2x)/(-x² - 3x +4) > 1 x+4+2-2x>-x²-3x+4 X²+3x-4+x+4+2-2x>0 x²+2x+2>0 D = 2² - 4 ·2 = 4 - 8 = -4 KORNEJ NETU
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solinjsh | Smotri fail! | | |
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mggt | 1) 2÷x-4 > 1÷x+5
2(x-4) x+5 ___ > ____ = 2x-4 > x+5 (x-4) (x+5)
x>9 atbilde. ___.\\\\\ x xE (9;+∞) E - pieder 9 2) 2x2+x-6 ______ < = 0 man liekas tu te kko nepareizi esi uzraxstiijis 3x+4
2x2+x-6 <= 3x-4 5x-6 <= 3x-4 2x<=2 x=1 atbilde ja jazime likne,tad ___.\\\\\\> x xE [1;+∞)
3) 1(x+4) 2(x-1) ______ + _____ > (1-x)(x+4) (1-x)(x+4) (1-x)(x+4) x+4+2x-2 > x+4-x²-4x x²+6x-2 > 0 D=36-4*1*(-2) D=√44 = √4*11 = 2√11 | |
| № 26145, Algebra, 10 klase Fotka estj pozalujsta pomogite , foto ir ludzu palidziet man :) | | |
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vespertilio | Skat. failu. | | |
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arieta | failaa | |
| № 26168, Algebra, 10 klase √3x²+7x-6lielaks vai vienads 0 | | |
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hhlady | √3x²+7x-6lielaks vai vienads 0 3x²+7x-6=0 a=3 b=7 c=-6 D=49+72=121 x1=-7+11 dalīts ar 6=viena trešdaļa x2=-7-11 dalīts ar 6=-3 D[y]=[no -3 līdz vienai trešdaļai ieskaitot] būs pareizi. :)
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marke | 3x^2 - 7x - 6=0 3x^2 - 9x + 2x - 6 = 0 3x(x - 3) + 2(x - 3) = 0 (3x + 2)(x - 3) = 0 vienads 3x + 2 = 0 vai x-3 = 0 x= -2/3 vai x=3 | |
| № 26402, Algebra, 10 klase atrisināt nevienādību | | |
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Kuziņš | X²-10x+25>0 D=(-10)2 -4*1*25=100-100=0
X1= 10+√D/2*1=10+0/2=10/2=5 x2= 10-√D/2*1=10-0/2=10/2=5
x pieder (-∞;+∞) | | |
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inesuxx | peec dotajiem a=1, b= -10, c= 25 10 ² -4 *1 *25=100-100=0 X1=10-√0 daliits ar divi = 5 x2 = 10+√0 daliits ar divi =5 paarbaude 5²-10*5 +25 =0 | | |
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snow | X²-10x+25>0 X² - 10x + 25 = 0 D = 100 - 4 · 1 · 25 = 100 - 100 = 0 x1,2 = 10 ÷ 2 = 5 x pieder (- bezgaliba; + bezgaliba)
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vespertilio | pielikumā | | |
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puuteejaXD | D=10²-4·1·25=0
tā kā D=O, tad ir tikai viena sakne
x=10÷2=5 | |
| | № 26424, Algebra, 10 klase 1istēma x2-y2=7 x2+y2=25 2sistēma x2+xy-y2=11 x-y=2 3sistēma x2-y2=5 xy=6 4sistēma x2+3x-4y=20 x2+6x+y=-5 | | |
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angel | failā | |
| № 26454, Algebra, 10 klase Pomogite plz reshitj Fotka estj :))) | | |
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arieta | v faile | |
| № 26507, Algebra, 10 klase 2x²+3x-14 (x-2)(x+3) i vse eto boljwe 0 | | |
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catwoman | 2x²+3x-14>0 2x²+3x-14=0 D=9+112=121 x=-3+11/4=2 x=-3-11/4=7 x pieder (-∞;2) apvienojuma (7;+∞)
(x-2)(x+3)>0 x-2>0 x+3>0 x>2 x>-3 x pieder (2;+∞) | | |
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angel | 2x²+3x-14>0 D=9+112=121 x>(-3+11)/4=2 x<(-3-11)/4=-7 x (-∞;-7)U(2;+∞)
(x-2)(x+3)>0 x>2 x<-3 x (-∞;-3)U(2;+∞) | | |
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Lacis | 1.2x²+3x-14= D=3²-4*2*(-14)=121 x1= -3+11:4=2 x2=-3-11:4=-3,5 2,(x-2)(x+3)= x²+3x-2x-6 x²+x-6=0 D=1-4*1*(-6)=1+24=25 x1= -1+5:2=2 x2= -1-5:2= -3 | | |
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Blizko | 2x²+3x-14=0; D=9+112=121 x1=(-3+11)/4=2 x2=(-3-11)/4=-3.5 Otvet: X prinadlezit ot -3,5 do 2 (ne vklju4aja oba zna4enija)
(x-2)(x+3)=0 x=2; x=-3 Otvet: X prinadlezit ot -3 do 2 (ne vklju4aja oba zna4enija)
Zelaju uda4i ;)
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snow | 2x²+3x -14>0 D = 3²-4*2*-14 = 9 + 112 = 121 X1 = (-3-11)/4 = -3,4 X2 = (-3+11)/4 = 2 xpridnalizhit(-∞;-3,4)U(2;+∞)
(x-2)(x+3) > 0 Budet dva vaianta 1) x-2>0 2) x-2<0 x+3>0 x+3<0 x>2 x<2 x>-3 x<-3 x(-∞;-3)U(2;+∞) | |
| № 26848, Algebra, 10 klase Pozalujsta pomogite reshitj foto estj ... | | |
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Zance18 | e) 3(5x+x²-ka pakape)=3(2*3-ka pakape) 5x+x²=6 x²+5x-6=0 D=25-4*(-6)=25+24=49 x1=(-5-7)/2=-6 x2=(-5+7)/2=1
18.uzd b 2/3(x pakape)=1,5 (pakape 1/4) 1,5( pakape -x)=1,5(pakape 1/4) -x=1/4 x=-1/4 19.a 3/4(pakape x)=3/4(pakape -x) x=-x x+x=o 2x=0 x=0
c) 3x-7=-(7x-3) 3x+7x=7+3 10x=10 x=1 G) 0.2* 5³/5(pakape x)=25(pakape x)/5²= (0.2*125)/5 pakape x=25 pakape x/5²= 25/5pakape x=25 pakape x/5² | |
| | № 26878, Algebra, 10 klase X³+2x²-9x-18=0 x³+6x²-9x-54=0
Rewitj neravenstvo! Uda4i :), eto legko, mne interesno u kovo mozg pawet i vam bonusi pojdut! | | |
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Nekit | x³+2x-9x-18=0 x(x²-9)+2(x²-9)=0 (x²-9)(x+2)=0 (x+3)(x-3)(x+2)=0 1) x=-3 2) x=3 3) x=-2
x³ +6x²-9x-54=0 x(x²-9)+6(x²-9)=0 (x+6)(x+3)(x-3)=0 1) x= -6 2) x= -3 3) x=3 | | |
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snow | X³+2x²-9x-18=0 x²(x+2) -9(x+2)=0 (x²-9)(x+2)=0 x²=9 x1 = 3 x2 = -3 x3 = -2
x³+6x²-9x-54=0 (x²-9)(x+6)=0 x1=3 x2 = -3 x = -6
P.S. Ja uchus toka vo 9. klase =D
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vespertilio | 1) x²(x+2)-9(x+2)=0 (x²-9)(x+2)=0 (x-3)(x+3)(x+2)=0 x1=3; x2=-3; x3=-2
2) x²(x+6)-9(x+6)=0 (x²-9)(x+6)=0 (x-3)(x+3)(x+6)=0 x1=3; x2=-3; x3=-6
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