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) = x^1/3

f(15) և  f(14)

f(15) = 15^1/3    f(14) = 14^1/3    f(15) > f(14)

4. Լուծել հավասարումը

ա) 2^x = 0,5

2^x = ½    2^x = 2^-1    x = -1

բ) ( 1/3)^x = ∛3

1/3 = 3^-1    3^-x = 3^1/3    -x = 1/3    x = -1/3

գ) 4^x-1 = 2×8^x-2

4 = 2²    8 = 2³    2^2x-2 = 2×2^3x-6

2x-2 = 1+3x-6    2x-3x = -6+2+1    -x = -3    x = 3

դ) 5^4x = (0,2)^x-6

0.2 = 5^-1    5^4x = 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 = 2^0.5    -9+3x = 0.5    3x = 9.5    x = 9.5/3

զ) (√0,5)^2-x = 32
(√0,5)^2-x = (0.5^0.5)^2-x = (2^-1)^1-0.5x = 2^0.5x-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) 3^x+1 × 5^x-2 < 27

3^x × 3 × 5^x × 1/25 < 27    15^x × 3/25 < 27    15^x < 225    15^x < 15²    x <  2<2 o:p="">

6. Հաշվել.

ա) ⤳ log₃81

3^x = 81    x = 4

⤳ lg0,001

10^x = 0.001    x = -3

⤳ log_1/749√7

(1/7)^x = 49√7    x = -2.5

⤳ 5^2log₅¹²

(5^2log₅¹²)² = 12² = 144

⤳ log₂₅1/125

25^x = 1/125    x = -1.5

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

⤳ 2log_1/56 - 1/2log_1/5400 - 4log_1/5∜45 =

= log_1/536 – log_1/520 – log_1/545 = log_1/50.04    1/5^x = 0.04    x = 2

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

⤳ 36^log₆⁵ + 10^1-lg2 - 3^log₉³⁶ =

⤳ (25^log_0,2⁶+4^log_0,5⁶)^1/lg18 =

7. Լուծել հավասարումը

⤳ log_0,9(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