Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 628

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 = 0 x² − 22x − 23 = 0 D = b² − 4ac = (−22)² − 4 ⋅ 1 ⋅ (−23) = 484 + 92 = 576 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{22+\sqrt{576}}{2 ⋅ 1} $$ = 22 + 24/2 = 46/2 = 23 x₂ = 22 − 24/2 = −2/2 = −1
x₁ = 23, x₂ = −1
б) x² + x/2 = 8x − 7/3 3(x² + x) = 2(8x − 7) 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 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{13+\sqrt{1}}{2 ⋅ 3} $$ = 13 + 1/6 = 14/6 = 7/6 = 11/6 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{13-\sqrt{1}}{2 ⋅ 3}= $$ = 13 − 1/6 = 12/6 = 2
x₁ = 11/6, x₂ = 2
в) x − 3 = 1 − x²/3 x − 3/1 = 1 − x²/3 3(x − 3) = 1 − x² 3x − 9 = 1 − x² 3x − 9 − 1 + x² = 0 x² + 3x − 10 = 0 D = b² − 4ac = 3² − 4 ⋅ 1 ⋅ (−10) = 9 + 40 = 49 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-3+\sqrt{49}}{2 ⋅ 1} $$ = −3 + 7/2 = 4/2 = 2 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{-3-\sqrt{49}}{2 ⋅ 1} $$ = −3 − 7/2 = −10/2 = −5
x₁ = 2, x₂ = −5
г) 2 − x²/7 = x/2 2(2 − x²) = 7x 4 − 2x² − 7x = 0 −2x² − 7x + 4 = 0 D = b² − 4ac = (−7)² − 4 ⋅ (−2) ⋅ 4 = 49 + 32 = 81 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{7+\sqrt{81}}{2(-2)}= $$ = 7 + 9/−4 = 16/−4 = −4 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{7-\sqrt{81}}{2(-2)} $$ = 7 − 9/−4 = −2/−4 = 0,5
x₁ = −4, x₂ = 0,5