17 May 2019

Տարեվերջյան ամփոփիչ աշխատանք


Տարեվերջյան ամփոփիչ աշխատանք

1. Բաղդատել թվերը, եթե
ա) f(x) = x8

f(3) և  f(5)
f(3) = 38    f(5) = 58    f(3) < f(5)

f(-11)  և  f(12)
f(-11) = -118    f(12) = 128    f(-11) < f(12)

բ) f(x) = x3

f(-2)  և  f(-7)
f(-2) = -23    f(-7) = -73    f(-2) > f(-7)

f(10) և f(12)
f(10) = 103    f(12) = 123    f(10) < f(12)


2.Լուծել հավասարումը

ա )x4=74
x = 7

3. Բաղդատել թվերը   f(x) = x1/3

f(15) և  f(14)
f(15) = 151/3    f(14) = 141/3    f(15) > f(14)


4. Լուծել հավասարումը

ա) 2x = 0,5

2x = ½    2x = 2-1    x = -1

բ) ( 1/3)x = ∛3
1/3 = 3-1    3-x = 31/3    -x = 1/3    x = -1/3

գ) 4x-1 = 2×8x-2
4 = 22    8 = 23    22x-2 = 2×23x-6
2x-2 = 1+3x-6    2x-3x = -6+2+1    -x = -3    x = 3

դ) 54x = (0,2)x-6
0.2 = 5-1    54x = 5-x+6    4x = -x+6    5x = 6    x = 1.2

ե) (0,125)3-x = 2 √2
0.125 = 2-3    2-9+3x = 20.5    -9+3x = 0.5    3x = 9.5    x = 9.5/3

զ) (√0,5)2-x = 32
(√0,5)2-x = (0.50.5)2-x = (2-1)1-0.5x = 20.5x-1    0.5x-1 = 5    0.5x = 6    x =12


5. Լուծել անհավասարումը
ա) (1/4)x ≤ 64
1/4 = 0.25    64 = 0.25-3    x ≤ -3

բ) 3x+1 × 5x-2 < 27
3x × 3 × 5x × 1/25 < 27    15x × 3/25 < 27    15x < 225    15x < 152    x <  2<2 o:p="">


6. Հաշվել.
ա) log381
3x = 81    x = 4

lg0,001
10x = 0.001    x = -3

log1/749√7
(1/7)x = 49√7    x = -2.5

52log512
(52log512)2 = 122 = 144

log251/125
25x = 1/125    x = -1.5

 բ) 2log26 -log29 = log236 – log29 = log24    2x = 4    x = 2

2log1/56 - 1/2log1/5400 - 4log1/5∜45 =
= log1/536 – log1/520 – log1/545 = log1/50.04    1/5x = 0.04    x = 2

(log536 – log512) ÷ log59 = log53 ÷ log59 = log93    9x = 3    x = 0.5

36log65 + 101-lg2 - 3log936 =

(25log0,26+4log0,56)1/lg18 =

7. Լուծել հավասարումը
log0,9(6x-23) = 0
6x – 23 = 0.90    6x – 23 = 1    6x = 24    x = 4

2log2(x-5) + log √2(x+2) = 6
2log2(x-5) + 2log2(x+2) = 6    2log2(x2-3x-10) = 6
log2(x2-3x-10) = 3    x2-3x-10 = 8    x2-3x-18 = 0    D = 9+72 = 81
x1 = (3+9)/2 = 6    x2 = (3-9)/2 = -3

1/ lg10x +6/ lgx+5 = 1
lg10x = lg10+lgx = 1+lgx    (lgx+5+6(lgx+1))/((lgx+1)(lgx+5)) = 1
(7lgx+11)/((lgx+1)(lgx+5)) = 1    7lgx+11 = lg2x+5lgx+lgx+5
lg2x-lgx-6 = 0    նշ․ lgx = t    t2+t+6 = 0    D = 1+24 = 25

x1 = (1+5)/2 = 3    x2 = (1-5)/2 = -2


8. Լուծել անհավասարումը

log2(x-5) ≥ 3
x-5 ≥ 23    x-5 ≥ 8    x ≥ 13

log3(x2+7x-5) < 1<1 o:p="">
x2+7x-5 < 3    x2+7x-8 < 3    D = 49+32 = 81    x1 = 2    x2 = -8

lg(7x+5) < 1+ lg3
lg(7x-5) – lg3 < 1    (7x+5)/3 < 10    7x+5 < 30    7x < 25    x < 25/7

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