与えられた13個の数式を計算し、簡単にすることを求められています。

1. 問題の内容

与えられた13個の数式を計算し、簡単にすることを求められています。

2. 解き方の手順

(1) $-5 + 7 = 2$

(2) $1 - (-3) = 1 + 3 = 4$

(3) $6 + (-2)^2 = 6 + 4 = 10$

(4) $2x \times (-x) = -2x^2$

(5) $\frac{2}{3}a + \frac{1}{4}a = \frac{8}{12}a + \frac{3}{12}a = \frac{11}{12}a$

(6) $6a^3b \times \frac{b}{3} \div 2a = \frac{6a^3b^2}{3} \div 2a = 2a^3b^2 \div 2a = \frac{2a^3b^2}{2a} = a^2b^2$

(7) $6xy^2 \div (-\frac{3}{5}xy) \div (-2x)^3 = 6xy^2 \times (-\frac{5}{3xy}) \div (-8x^3) = -\frac{30xy^2}{3xy} \div (-8x^3) = -10y \div (-8x^3) = \frac{-10y}{-8x^3} = \frac{5y}{4x^3}$

(8) $\frac{8a+9}{4} - \frac{6a+4}{3} = \frac{3(8a+9) - 4(6a+4)}{12} = \frac{24a+27-24a-16}{12} = \frac{11}{12}$

(9) $(x+5)(x+4) = x^2 + 4x + 5x + 20 = x^2 + 9x + 20$

(10) $(x-3y)(x+3y) = x^2 - (3y)^2 = x^2 - 9y^2$

(11) $(x+1)(x-1) - (x+3)(x-8) = (x^2 - 1) - (x^2 - 8x + 3x - 24) = (x^2 - 1) - (x^2 - 5x - 24) = x^2 - 1 - x^2 + 5x + 24 = 5x + 23$

(12) $(2x-3)^2 - 4x(x-1) = (4x^2 - 12x + 9) - (4x^2 - 4x) = 4x^2 - 12x + 9 - 4x^2 + 4x = -8x + 9$

(13) $(3x-4y)(5x+y) - 3(2x-3y)^2 - (x-2y)(x+2y) = (15x^2 + 3xy - 20xy - 4y^2) - 3(4x^2 - 12xy + 9y^2) - (x^2 - 4y^2) = (15x^2 - 17xy - 4y^2) - (12x^2 - 36xy + 27y^2) - (x^2 - 4y^2) = 15x^2 - 17xy - 4y^2 - 12x^2 + 36xy - 27y^2 - x^2 + 4y^2 = (15-12-1)x^2 + (-17+36)xy + (-4-27+4)y^2 = 2x^2 + 19xy - 27y^2$

3. 最終的な答え

(1) 2

(2) 4

(3) 10

(4) $-2x^2$

(5) $\frac{11}{12}a$

(6) $a^2b^2$

(7) $\frac{5y}{4x^3}$

(8) $\frac{11}{12}$

(9) $x^2 + 9x + 20$

(10) $x^2 - 9y^2$

(11) $5x + 23$

(12) $-8x + 9$

(13) $2x^2 + 19xy - 27y^2$