Question

№ 5.38 № 5.39 № 540 № 541​

Answer 1

Ответ:

$5.38 \\ {x}^{2} - {y}^{2} = (x - y)(x + y) \\ {m}^{2} - {n}^{2} = (m - n)(m + n) \\ {c}^{2} - 25 = (c - 5)(c + 5) \\ {a}^{2} - 1 = (a - 1)(a + 1) \\ 25 - {a}^{2} = (5 - a)(5 + a) \\ 49 - {b}^{2} = (7 - b)(7 + b) \\ 100 - {p}^{2} = (10 - p)(10 + p) \\ {m}^{2} - 400 = (m - 20)(m + 20) \\ {b}^{2} - 0.04 = (b - 0.2)(b + 0.2) \\ 1.21 - {x}^{2} = (1.1 - x)(1.1 + x) \\ {n}^{2} - \frac{4}{9} = (n - \frac{2}{3} )(n + \frac{2}{3} ) \\ \frac{25}{64} - {p}^{2} = ( \frac{5}{8} - p)( \frac{5}{8} + p)$

$5.39 \\ 9 {a}^{2} - 25 {b}^{2} = (3a - 5b)(3a + 5b) \\ 4 {c}^{2} - 49 {d}^{2} = (2c - 7d)(2c + 7d) \\ - 81 + 25 {m}^{2} = ( - 9 + 5m)( 9 + 5m) \\ {x}^{2} {y}^{2} - 0.04 = (xy - 0.2)(xy + 0.2) \\ 0.16 - {x}^{2} = (0.4 - x)(0.4 + x) \\ 144 - 49 {n}^{2} = (12 - 7n)(12 + 7n) \\ {a}^{2} {b}^{2} - {c}^{2} = (ab - c)(ab + c) \\ {p}^{2} {q}^{2} - 4 {k}^{2} = (pq - 2k)(pq + 2k)$

$5.40 \\ (30 + 1)(30 - 1) = {30}^{2} - 1 = 900 - 1 = 899 \\ 61 \times 59 = (60 + 1)(60 - 1) = {60}^{2} - 1 = 3600 - 1 = 3599 \\ 199 \times 201 = {200}^{2} - 1 = 40000 - 1 = 39999 \\ 72 \times 68 = {70}^{2} - {2}^{2} = 4900 - 4 = 4896 \\ {55}^{2} - {45}^{2} = 10 \times 100 = 1000 \\ {41}^{2} - {31}^{2} = 10 \times 72 = 720 \\ {76}^{2} - {24}^{2} = 52 \times 100 = 5200 \\ {37}^{2} - {23}^{2} = 14 \times 60 = 840$

$5.41 \\ {x}^{2} - 9 = 0 \\ (x - 3)(x + 3) = 0 \\ x1 = 3 \: \: \: x2 = - 3 \\ \\ {x}^{2} - 0.04 = 0 \\ (x - 0.2)(x + 0.2) = 0 \\ x1 = 0.2 \: \: \: x2 = - 0.2 \\ \\ {x}^{2} - 81 = 0 \\ (x - 9)(x + 9) = 0 \\ x1 = 9 \: \: \: x2 = - 9 \\ \\ {x}^{2} - \frac{1}{9} = 0 \\ (x - \frac{1}{3} )(x + \frac{1}{3} ) = 0 \\ x1 = \frac{1}{3} \: \: \: x2 = - \frac{1}{3} \\ \\ {y}^{2} - 1 \frac{9}{16} = 0 \\ (y - \frac{5}{4} )(y + \frac{5}{4} ) = 0 \\ y1 = 1 \frac{1}{4} \: \: \: y2 = - 1 \frac{1}{4} \\ \\ {y}^{2} - 2 \frac{1}{4} = 0 \\ (y - \frac{3}{2} )(y + \frac{3}{2} ) = 0 \\ y1 = 1 \frac{1}{2} \: \: \: y2 = - 1 \frac{1}{2}$