Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 624

а) 4x + 4/3x² + 2x − 1 3x² + 2x − 1 = 0 D = b² − 4ac = 2² − 4 ⋅ 3 ⋅ (−1) = 4 + 12 = 16 > 0 имеет 2 корня x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-2+\sqrt{16}}{2 ⋅ 3}= $$ = −2 + 4/6 = 2/6 = 1/3 x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-2-\sqrt{16}}{2 ⋅ 3} $$ = −2 − 4/6 = −6/6 = −1 x₁ = 1/3, x₂ = −1 3x² + 2x − 1 = 3(х − 1/3)(х + 1) = (3х − 1)(х + 1) 4x + 4/3x² + 2x − 1 = 4(x + 1)/(3x − 1)(x + 1) = 4(x + 1) : (x + 1)/(3x − 1)(x + 1) : (x + 1) = 4/3x − 1
б) 2a² − 5a − 3/3a − 9 2а² − 5а − 3 = 0 D = b² − 4ac = (-5)² − 4 ⋅ 2 ⋅ (−3) = 25 + 24 = 49 > 0 имеет 2 корня a₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{5+\sqrt{49}}{2 ⋅ 2} $$ = 5 + 7/4 = 12/4 = 3 a₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{5-\sqrt{49}}{2 ⋅ 2} $$ = 5 − 7/4 = −2/4 = −0,5 a₁ = 3, a₂ = −0,5 2а² − 5а − 3 = 2(а − 3)(а + 0,5) = (а − 3)(2а + 1) 2a² − 5a − 3/3a − 9 = (a − 3)(2a + 1)/3(a − 3) = (a − 3)(2a + 1) : (a − 3)/3(a − 3) : (a − 3) = 2a + 1/3
в) 16 − b²/b² − b − 12 b² − b − 12 = 0 D = b² − 4ac = (−1)² − 4 ⋅ 1 ⋅ (−12) = 1 + 48 = 49 > 0 имеет 2 корня b₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{1+\sqrt{49}}{2 ⋅ 1} $$ = 1 + 7/2 = 8/2 = 4 b₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{1-\sqrt{49}}{2 ⋅ 1} $$ = 1 − 7/2 = −6/2 = −3 b₁ = 4, b₂ = −3 b² − b − 12 = (b − 4)(b + 3) 16 − b²/b² − b − 12 = (4 − b)(4 + b)/(b − 4)(b + 3) = −(b − 4)(4 + b) : (b − 4)/(b − 4)(b + 3) : (b − 4) = −(4 + b)/b + 3
г) 2y² + 7y + 3/y² − 9 2y² + 7y + 3 = 0 D = b² − 4ac = 7² − 4 ⋅ 2 ⋅ 3 = 49 − 24 = 25 > 0 имеет 2 корня y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-7+\sqrt{25}}{2 ⋅ 2} $$ = −7 + 5/4 = −2/4 = −0,5 y₂ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{-7-\sqrt{25}}{2 ⋅ 2} $$ = −7 − 5/4 = −12/4 = −3 y₁ = −0,5, y₂ = −3 2y² + 7y + 3 = 2(y + 0,5)(y + 3) = (2y + 1)(y + 3)
д) p² − 11p + 10/20 + 8p − p² p² − 11p + 10 = 0 D = b² − 4ac = (−11)² − 4 ⋅ 1 ⋅ 10 = 121 − 40 = 81 > 0 имеет 2 корня p₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{11+\sqrt{81}}{2 ⋅ 1} $$ = 11 + 9/2 = 20/2 = 10 p₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{11-\sqrt{81}}{2 ⋅ 1} $$ = 11 − 9/2 = −2/2 = −1 p₁ = 10, p₂ = −1 p² − 11p + 10 = (p − 10)(p + 1)
−p² + 8p + 20 = 0 D = b² − 4ac = 8² − 4 ⋅ (−1) ⋅ 20 = 64 + 80 = 144 > 0 имеет 2 корня p₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-8+\sqrt{144}}{2(-1)} $$ = −8 + 12/−2 = 4/−2 = −2 p₂ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{-8-\sqrt{144}}{2(-1)} $$ = −8 − 12/−2 = −20/−2 = −10 p₁ = −2, p₂ = −10 −p² + 8p + 20 = −(p + 2)(p − 10) p² − 11p + 10/20 + 8p − p² = (p − 10)(p + 1)/−(p + 2)(p − 10) = (p − 10)(p + 1) : (p − 10)/−(p + 2)(p − 10) : (p − 10) = p + 1/−(p + 2)
е) 3x² + 16x − 12/10 − 13x − 3x² 3x² + 16x − 12 = 0 D = b² − 4ac = 16² − 4 ⋅ 3 ⋅ (−12) = 256 + 144 = 400 > 0 имеет 2 корня x₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{-16+\sqrt{400}}{2 ⋅ 3} $$ = −16 + 20/6 = 4/6 = 2/3 x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-16-\sqrt{400}}{2 ⋅ 3} $$ = −16 − 20/6 = −36/6 = −6 x₁ = 2/3, x₂ = −6 3x² + 16x − 12 = 3(х − 2/3)(х + 6) = (3х − 2)(х + 6)
−3x² − 13x + 10 = 0 D = b² − 4ac = (−13)² − 4 ⋅ (−3) ⋅ 10 = 169 + 120 = 289 > 0 имеет 2 корня x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{13+\sqrt{289}}{2(-3)} $$ = 13 + 17/−6 = 30/−6 = −5 x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{13-\sqrt{289}}{2(-3)} $$ = 13 − 17/−6 = −4/−6 = 2/3 x₁ = −5, x₂ = 2/3 10 − 13х − 3х² = −3(х + 5)(х − 2/3) = −(х + 5)(3х − 2)
3x² + 16x − 12/10 − 13x − 3x² = (3x − 2)(x + 6)/−(x + 5)(3x − 2) = (3x − 2)(x + 6) : (3x − 2)/−(x + 5)(3x − 2) : (3x − 2) = x + 6/−(x + 5)