Question

25 БАЛЛОВ ТОМУ КТО ПОДРОБНО РЕШИТ 3,4,5 ЗАДАНИЕ..

Answer 1

3. (x² - 2x)² - 7 (x² - 2x) - 8 = 0

t² - 7t - 8 = 0

t = 8

t = -1

x² - 2x = 8

x² - 2x = -1

x = 4

x = -2

x = 1

x₁ = -2, x₂ = 1, x₃ = 4

4. (x⁴ - 3x + 1) * (x² - 3x + 3) = 3

x⁴ = 3x³ + 3x² - 3x³ + 9x² - 9x + x² -3x + 3 = 3

x⁴ = 3x³ + 3x² - 3x³ + 9x² - 9x + x² -3x = 0

x⁴ - 6x³ + 13x² - 12x = 0

x * (x³ - 6x² + 13x - 12) = 0

x * (x³ - 3x² - 3x² + 9x + 4x - 12) = 0

x * (x² * (x-3) - 3x * (x-3) + 4 (x-3)) = 0

x * (x-3) * (x² - 3x + 4) = 0

x = 0

x - 3 = 0

x² - 3x + 4 = 0

x = 0

x = 3

x ⊄ R

x = 0

x = 3

x₁ = 0, x₂ = 3

5. (7x-2)² - 12 * I7x - 2I - 28 = 0

t² - 12 * ItI - 28 = 0

t  = 14

t = -14

7x - 2 = 14

7x - 2 = -14

x = $\frac{16}{7}$

x = $-\frac{12}{7}$

x₁ = $-\frac{12}{7}$, x₂ = $\frac{16}{7}$

Answer 2

3)
(х²-2х)=t
t²-7t-8=0
d=7²+4*8=49+32=91
√81=9
t1=(7+9/2)=16/2=8
t2=(7-9/2)=-2/2=-1
4)
(x²-3x+1)(x²-3x+3)=3
x⁴-3x³+3x²-3x³+9x²-9x+x²-3x+3=x⁴-6x³+13x²-12x=0
x(x³-6x²+13x-12)=0
x(x³-3x²-3x²+9x+4x-12)=0
x(x²(x-3)-3x(x-3)+4(x-3))=0
x(x-3)(x²-3x+4)=0
x=0
x-3=0
x²-3x+4=0
x=3
5)
t²-12|t|-28=0
t²-12t-28=0
t²-12(-t)-28=0
d=12²+4*28=144+112=256
√256=16
t=(12-16/2)=-4/2=-2
t=(12+16/2)=28/2=14