ЛЕКЦІЇ ² ВПРАВИ

14.2.4. Êîðåí³ òà ¿õí³ âëàñòèâîñò³:
( )
m
n
n m a a
ab = a b, a ≥0, b ≥ 0; a = a , a ≥ 0; = a ≥ 0, b > 0;
n
b b
n n n n n
a = a a = a a = a a ≥ .
n m nm nk mk n m 2n
2n
; ; , 0
( n a — àðèôìåòè÷íå çíà÷åííÿ êîðåíÿ ïðè a ≥ 0).
14.2.5. Ëîãàðèôìè òà ¿õí³ âëàñòèâîñò³:
( )
log b = c ⇔ a c
= b a a
> 0; a ≠ 1; b > 0 ;
10
( )
lg b= log b; ln b= log b, b> 0 e ≈2,718
;
( )
loga b
a = b; loga bc = loga b + log
a c , bc > 0 ;
b
n 1
loga = loga b − log
a
c , bc > 0; loga b = log
a
b, b > 0 ;
c
n
e
( )
log b α =α log b b > 0, α∈R ;
a
a
logc
b
log;
ab = 1
log
ab ( b , a , b , a )
loglog
a
= a
≠ 1 ≠ 1 > 0 > 0 ;
b
c
( )
α
log a = 1; log α b = log b α∈R , α ≠ 0, a > 0, a ≠ 1, b > 0 ;
a
a
a
( )
log,
a
1= 0 a > 0 a≠1 ;
1 1
= lg, e = 0 43429K; = ln 10 = 2,
30258K ln10
lg e
.
14.2.6. Êîðåí³ êâàäðàòíîãî ð³âíÿííÿ. Ðîçêëàä êâàäðàòíîãî
òðè÷ëåíà íà ìíîæíèêè:
2
− b±
D
2
ax + bx + c = 0 ( a ≠0 ), x12 ,
= , D = b −4ac
≥0;
2a
2
− k±
D
2
ax + 2kx + c = 0 ( a ≠ 0 ), x12 ,
= , D = k −ac
≥0 ;
a
2
p D
2
x + px+ q = 0 , x12 ,
=− ± , D = b −4ac≥0.
2 2
Ôîðìóëè ³ºòà:
2
x + px+ q = 0 , xx = q,
x + x = −p
Á³êâàäðàòíå ð³âíÿííÿ
1 2 1 2
.
( )
4 2
ax + bx + c = 0 a ≠0
ï³äñòàíîâêîþ x 2 = y çâîäèòüñÿ äî êâàäðàòíîãî ð³âíÿííÿ
( )
+ + = ≠
2
ay by c 0 a 0 .
Ðîçêëàä òðè÷ëåíà íà ìíîæíèêè:
( )( )
2
ax + bx + c = a x −x x −x
1 2 ,
äå x 1 ³ x 2 — êîðåí³ êâàäðàòíîãî ð³âíÿííÿ.
14.2.7. Àðèôìåòè÷íà ïðîãðåñ³ÿ
a
a , a , a , K 1 2 3
, a n
, K;
an = a1+ d( n− 1);
d = a2 − a1 = K= an − an
− 1
= K;
a
+ a
2
( n 2)
n− 1 n+
1
n
= ≥
;
n n
a1+ a = a2 + a − 1
= K ;
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