Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 544

а) x² − 1/2 − 11x = 11 x² − 1/2 − 11x ⋅ 2/1 ⋅ 2 = 11 ⋅ 2/1 ⋅ 2 x² − 1/2 − 22x/2 = 22/2 x² − 1 − 22x = 22 x² − 1 − 22x − 22 = 0 x² − 22x − 23 = 0 D = b² − 4ac = (−22)² − 4 ⋅ 1 ⋅ (−23) = 484 + 92 = 576 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{22+\sqrt{576}}{2 ⋅ 1} $$ = 22 + 24/2 = 46/2 = 23
x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{22-\sqrt{576}}{2 ⋅ 1} $$ = 22 − 24/2 = −2/2 = −1
x₁ = 23, x₂ = −1
б) x² + x/2 = 8x − 7/3 (x² + x) ⋅ 3 = (8x − 7) ⋅ 2 3x² + 3x = 16x − 14 3x² + 3x − 16x + 14 = 0 3x² − 13x + 14 = 0 D = b² − 4ac = (−13)² − 4 ⋅ 3 ⋅ 14 = 169 − 168 = 1 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{13+\sqrt{1}}{2 ⋅ 3} $$ = 13 + 1/6 = 14/6 = 7/3 = 21/3
x₂ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{13-\sqrt{1}}{2 ⋅ 3} $$ = 13 − 1/6 = 12/6 = 2
x₁ = 21/3, x₂ = 2
в) 4x² − 1/3 = x(10x − 9) 4x² − 1/3 = 10x² − 9x/1 (4x² − 1) ⋅ 1 = (10x² − 9x) ⋅ 3 4x² − 1 = 30x² − 27x 4x² − 1 − 30x² + 27x = 0 D = b² − 4ac = 27² − 4 ⋅ (−26) ⋅ (−1) = 729 − 104 = 625 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-27+\sqrt{625}}{2(-26)} $$ = −27 + 25/−52 = −2/−52 = 1/26
x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-27-\sqrt{625}}{2(-26)} $$ = −27 − 25/−52 = −52/−52 = 1
x₁ = 1/26, x₂ = 1
г) 3/4x² − 2/5x = 4/5x² + 3/4 3 ⋅ 5/4 ⋅ 5x² − 2 ⋅ 4/5 ⋅ 4x = 4 ⋅ 4/5 ⋅ 4x² + 3 ⋅ 5/4 ⋅ 5 15/20x² − 8/20x − 16/20x² − 15/20 = 0 −1/20x² − 8/20x − 16/20x² = 0 −x² − 8x − 15 = 0 D = b² − 4ac = (−8)² − 4 ⋅ (−1) ⋅ (−15) = 64 - 60 = 4 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{8+\sqrt{4}}{2(-1)}= $$ = 8 + 2/−2 = 10/−2 = −5
x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{8-\sqrt{4}}{2 \sqsubset(-1)} $$ = 8 − 2/−2 = 6/−2 = −3
x₁ = −5, x₂ = −3