数学
高校生

数1の展開

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3 次の式を展開せよ。 (1) (a²-2bc)(bc+3a²) *(3) (x−y)²(x+y)²(x² + y²)² (5) (x²+xy+y²)(x² −xy+y²) (7) (3x-y+1)(2x+y-1) (2) (m²-2m-1)² *(4) (a+b-c-d)(a-b-c+d) *(6) (k+2)(k-1) (k²-k+2) (8) (2a-2b+c)(a-b-c)
(2式) -y+3)+-y+7 別解 -2x²-2x-x+3x²+9y = 2x²7x+xy-3+8y+3 (与式)=2x+5-32x2(x2+5_3x) 21 (1) (5=(2x+(a=11² 4+5-3.x) 3x³ +15x9x²-2x¹10x² +6x³ =-2x+9x3-2327x20 x) = (2x) + a²+(-1)² +-y+3) 22の頃は +x-x = (2x)²2-2x (a-1a-1)² = 4x²a-4)x+ 2a + 1) +2-2x a+2.1)+2..(2x) =4²+ a² +1+x-2a- =4²+(4a-x+(a²-201) (2 (5)=a-2a-2-3+(ax-2-2).(-) 3ax-6a-6-ax² +20+2x x 3xxx(-x²)-3x³- -ax² +50+2)x+6) よって,xの係数は 4x²-20+ 12x =2x3 2 23 (1) (t)=(a²-2bc)(3a² + bc) (2) (与式)={(m² -2m-1}2 = 1·3(a ²)² + (1-bc+(-2bc)-3a² +(-2bc). bc = 3a4-5a²bc-2b²c² = (m²-2m)² -2.(m²-2m) 1+1² = (m²)²-2.m².2m +(2m)² -2m² +4m +1., =m¹-4m³ +4m² -2m² + 4m+i =m¹-4m³ +2m² +4m +1 SU (5x) = (m²)² + (−2m)² + (−1)² +2 m².(-2m) +2.(-2m) (-1) +2.(-1).m² =m¹ +4m² +1-4m³ +4m-2m² =m¹-4m³ +2m² +4m+1 (3) (5x) = {(x−y)(x+y)}² ×(x² + y²)² = (x² - y²)²(x² + y²) ² = {(x² - y²)(x² + y²)}² = {(x²)²-(y²)²² = (x² - y¹)² = (x¹)²-2-x²-y¹ + (y¹)² =x²-2x¹v¹ + po (4) (5x)=((a−c)+(b-d)||(a-c)-(b-d)) =(a-c)²-(b-d)² =a²-2ac+c² − (b² − 2bd+d²) =a²-b² +c²-d²-2ac+2bd (5) (5x) = {(x² + y²) + xy}{(x² + y²) − xy) = (x² + y²)²-(xy)² = ((x²)² +2+x²- y² + (y²)²) — x²y² = (x¹+2x²y² + y²¹)-x²y² = x² + x²y² + y² (6) (与式)=(k+2)(k-1)×(k²-k+2) =(k²+k-2)(k-k+2) = (k² + (k − 2)}{k²—(k-2)] =(k2)²-(k-2² = k¹-(k²-4k+4) = k¹-k² +4k-4 (7) (与式) = (3x-(y-1)}{2x+(y-1)) =3.2x2+(3.(y-1)-(y-1) ・2)x-(y-1)2 =6x²+(y-1)x-(y2-2y+1) =6x2+xy-x-y2+2y-1 =6x2+xy-y2-x+2y-1 (8) (与式) = 12(a-b+c)(a-b)-c) = 2.1 (a−b)²+(2-(-c)+c-1} (a - b) +c. (c) =2(a²-2ab+b²)-c(a−b)-c² =2a²-4ab+26²-ac+bc-c² = 2² +26²-c²-4ab+bc-ca 24 (1) (x)= x(x+3)x(x+1)(x+2) 3x)(x² + 2) x²+3x+3x)+2) =x2+3+2(x2+3 =x+x³ +9x² + +6x +6x³+1 +6x (2与式) (+1)(x-4) X-1)(x-2) (x²-3x4x²-3x+2 = [(x²-3-4}{(x² - 2x==2} =x²-3x)²-2(x²)-8 -5 4-6x³+9 2x²+6x-8 =x¹-6x³-7²+6x-8 (3)-¹)=(x+2x2)x(x+5)(-5) =(x-4)(x²-25) 数学Ⅰ TRIAL A'B'
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