Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 639

а) 10/(x − 5)(x + 1) + x/x + 1 = 3/x − 5 ОДЗ: x − 5 ≠ 0, x ≠ 5, x + 1 ≠ 0, x ≠ −1 10/(x − 5)(x + 1) + x(x − 5)/(x + 1)(x − 5) = 3(x + 1)/(x − 5)(x + 1) 10 + x² − 5x − 3x − 3 = 0 x² − 8x + 7 = 0 D = b² − 4ac = (−8)² − 4 ⋅ 1 ⋅ 7 = 64 − 28 = 36 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{8+\sqrt{36}}{2 ⋅ 1} $$ = 8 + 6/2 = 14/2 = 7 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{8-\sqrt{36}}{2 ⋅ 1} $$ = 8 − 6/2 = 2/2 = 1
x₁ = 7, x₂ = 1
б) 17/(x − 3)(x + 4) − 1/x − 3 = x/x + 4 ОДЗ: x − 3 ≠ 0, x ≠ 3, x + 4 ≠ 0, x ≠ −4 17/(x − 3)(x + 4) − 1(x + 4)/(x − 3)(x + 4) = x(x − 3)/(x + 4)(x − 3) 17 − x − 4 − x² + 3x = 0 −x² + 2x + 13 = 0 (−1) x² − 2x − 13 = 0 D = b² − 4ac = (−2)² − 4 ⋅ 1 ⋅ (−13) = 4 + 52 = 56 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{2+\sqrt{56}}{2 ⋅ 1}=\frac{2+2 \sqrt{14}}{2} $$ = 1 + $$\sqrt{14}$$ x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{2-\sqrt{56}}{2 ⋅ 1}=\frac{2-2 \sqrt{14}}{2} $$ = 1 − $$\sqrt{14}$$
x₁ = 1 + $$\sqrt{14}$$, x₂ = 1 − $$\sqrt{14}$$
в) 4/(x + 1)² − 1/(x − 1)² + 1/x² − 1 = 0 ОДЗ: x + 1 ≠ 0, x ≠ −1, x − 1 ≠ 0, x ≠ 1 4(x − 1)²/(x + 1)²(x − 1)² − 1(x + 1)²/(x − 1)²(x + 1)² + 1(x − 1)(x + 1)/(x − 1)(x + 1)(x − 1)(x + 1) = 0 4x² − 8x + 4 − x² − 2x − 1 + x² − 1 = 0 4x² − 10x + 2 = 0 : 2 2x² − 5x + 1 = 0 D = b² − 4ac = (−5)² − 4 ⋅ 2 ⋅ 1 = 25 − 8 = 17 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{5+\sqrt{17}}{2 ⋅ 2}=\frac{5+\sqrt{17}}{4} $$ x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{5+\sqrt{17}}{2 ⋅ 2}=\frac{5+\sqrt{17}}{4} $$
x₁ = $$ \frac{5+\sqrt{17}}{4} $$, x₂ = $$ =\frac{5-\sqrt{17}}{4} $$
г) 4/9x² − 1 + 1/3x² − x = 4/9x² − 6x + 1
D = b² − 4ac = (−8)² − 4 ⋅ 9 ⋅ (−1) = 64 + 36 = 100 > 0 имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{8+\sqrt{100}}{2 ⋅ 9} $$ = 8 + 10/18 = 18/18 = 1 x₂ = $$ \frac{-b-\sqrt{D}}{2 a}=\frac{8-\sqrt{100}}{2 ⋅ 9} $$ = 8 − 10/18 = −2/18 = −1/9
x₁ = 1, x₂ = −1/9