Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 326

а) х⁴ − 20х² + 64

х⁴ – 20х² + 64 = 0

Пусть t = x²
t² – 20t + 64 = 0

По теореме Виета:
t₁ = 16
t₂ = 4

х⁴ – 20х² + 64 = (x² – 16)(x² – 4) = (х – 4)(х + 4)(х – 2)(х + 2)

х⁴ – 20х² + 64 = (х – 4)(х + 4)(х – 2)(х + 2)

б) х⁴ − 17х² + 16

х⁴ – 17х² + 16 = 0

Пусть t = х²
t² – 17t + 16 = 0

По теореме Виета:
t₁ = 16
t₂ = 1

t² – 17t + 16 = (х² – 16)(х² – 1) = (х – 4)(х + 4)(х – 1)(х + 1)

х⁴ – 17х² + 16 = (х – 4)(х + 4)(х – 1)(х + 1)

в) х⁴ − 5х² + 36

х⁴ – 5х² – 36 = 0

Пусть t = х²
t² – 5y – 36 = 0

По теореме Виета
t₁ = 9
t₂ = –4

х⁴ – 20х² + 64 = ( х² – 9 )( х² + 4 ) = ( х – 3 )( х + 3 )( х² + 4)

х⁴ – 20х² + 64 = (х – 3)(х + 3)( х² + 4 )

г) х⁴ − 3х² − 4

х⁴ – 3х² – 4 = 0

Пусть t = х²
t² – 3t – 4 = 0

По теореме Виета
t₁ = –1
t₂ = 4

х⁴ – 3х² – 4 = (х² + 1)(х² – 4) = (х² + 1)(х – 2)(х + 2)

х⁴ – 3х² – 4 = (х² + 1)(х – 2)(х + 2)

д) 9х⁴ − 10х² + 1

9х⁴ – 10х² + 1 = 0

Пусть t = х²

9t² – 10t + 1 = 0

D₁ = (–5)² – 9 · 1 = 16

t₁ = $$\frac{5 + \sqrt{16}}{9}$$ = 5 + 4/9 = 9/9 = 1

t₂ = $$\frac{5 - \sqrt{16}}{9}$$ = 5 – 4/9 = 1/9

9х⁴ – 10х² + 1 = 9(х² – 1)(х² – 1/9) = 3 · 3(х – 1)(х + 1)(х – 1/3)(х + 1/3) = (х – 1)(х + 1)(3х – 1)(3х + 1)

9х⁴ – 10х² + 1 = (х – 1)(х + 1)(3х – 1)(3х + 1)

е) 4х⁴ − 17х² + 4

4х⁴ – 17х² + 4 = 0

Пусть t = x²

4t² – 17t + 4 = 0

D₁ = (–7)² – 4 · 4 · 4 = 289 – 64 = 2250

t₁ = $$\frac{17 + \sqrt{225}}{2 \cdot 4}$$ = 17 + 15/8 = 32/8 = 4

t₂ = $$\frac{17 - \sqrt{225}}{2 \cdot 4}$$ = 17 – 15/8 = 2/8 = 1/4

4х⁴ – 17х² + 4 = 4(х² – 1/4)(х² – 4) = 2 · 2 (х – 1/2)(х + 1/2)(х – 2)(х + 2) = (2х – 1)(2х +)(х – 2)(х + 2)

4х⁴ – 17х² + 4 = (2х – 1)(2х + 1)(х – 2)(х + 2)