# 数学演習/中学校3年生/式の計算/解答

## 展開(1)

1. ${\displaystyle z(x+y)=xz+yz}$
2. ${\displaystyle (x+4)(x+5)=x^{2}+5x+4x+20=x^{2}+9x+20}$
3. ${\displaystyle (x+3)^{2}=(x+3)(x+3)=x^{2}+3x+3x+9=x^{2}+6x+9}$
4. ${\displaystyle (x-6)^{2}=(x-6)(x-6)=x^{2}-6x-6x+36=x^{2}-12x+36}$
5. ${\displaystyle (x+5)(x-5)=x^{2}+5x-5x-25=x^{2}-25}$
6. ${\displaystyle (3x+4y)^{2}=(3x+4y)(3x+4y)=9x^{2}+12xy+12xy+16y^{2}=9x^{2}+24xy+16y^{2}}$
7. ${\displaystyle (x+2y)(-x+2y)=(2y+x)(2y-x)=-x^{2}+4y^{2}}$
8. ${\displaystyle 2(x-2y)^{2}=2(x-2y)(x-2y)=2(x^{2}-4xy+4y^{2})=2x^{2}-8xy+8y^{2}}$

## 展開(2)

1.解答の途中で${\displaystyle a^{2}+b^{2}=A}$としている。

{\displaystyle {\begin{aligned}(a+b)^{4}&=((a+b)^{2})^{2}\\&=(a^{2}+2ab+b^{2})^{2}\\&=(A+2ab)^{2}\\&=A^{2}+2\times A\times 2ab+4a^{2}b^{2}\\&=A^{2}+4abA+4a^{2}b^{2}\\&=(a^{2}+b^{2})^{2}+4ab(a^{2}+b^{2})+4a^{2}b^{2}\\&=(a^{4}+2a^{2}b^{2}+b^{4})+4a^{3}b+4ab^{3}+4a^{2}b^{2}\\&=a^{4}+4a^{3}b+(4+2)a^{2}b^{2}+4ab^{3}+b^{4}\\&=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}\\\end{aligned}}}

2.問題の${\displaystyle a+b=B}$とする。

{\displaystyle {\begin{aligned}(a+b+c)(a+b-c)&=(B+c)(B-c)\\&=B^{2}-c^{2}\\&=(a+b)^{2}-c^{2}\\&=(a^{2}+2ab+b^{2})-c^{2}\\&=a^{2}+2ab+b^{2}-c^{2}\\\end{aligned}}}

## 因数分解(1)

1. ${\displaystyle xy+yz=y(x+z)}$
2. ${\displaystyle x^{2}+5x+6=(x+2)(x+3)}$
3. ${\displaystyle x^{2}+10x+25=(x+5)^{2}}$
4. ${\displaystyle x^{2}-8x+16=(x-4)^{2}}$
5. ${\displaystyle x^{2}-100=(x+10)(x-10)}$
6. ${\displaystyle 12x^{2}+x-1=(3x+1)(4x-1)}$
7. ${\displaystyle 9x^{2}-4y^{2}=(3x+2y)(3x-2y)}$
8. ${\displaystyle 3x^{2}+12x+12=3(x^{2}+4x+4)=3(x+2)^{2}}$
9. ${\displaystyle (4x+9)(x+4)-11=4x^{2}+25x+36-11=4x^{2}+25x+25=(4x+5)(x+5)}$

## 因数分解(2)

1.問題の${\displaystyle x^{3}=X}$とする。

{\displaystyle {\begin{aligned}x^{6}-x^{3}-6&=(x^{3})^{2}-x^{3}-6\\&=X^{2}-X-6\\&=(X+2)(X-3)\\&=(x^{3}+2)(x^{3}-3)\\\end{aligned}}}

2.問題の${\displaystyle 2x+3y=C,3x-4y+D}$とする。

{\displaystyle {\begin{aligned}(2x+3y)^{2}-(3x-4y)^{2}&=C^{2}-D^{2}\\&=(C+D)(C-D)\\&=(2x+3y+3x-4y)(2x+3y-3x+4y)\\&=(5x-y)(-x+7y)\\\end{aligned}}}

## 展開と因数分解の利用

1. ${\displaystyle 201\times 199=(200+1)(200-1)=200^{2}-1^{2}=39999}$
2. ${\displaystyle 105^{2}=(100+5)^{2}=100^{2}+2\times 100\times 5+5^{2}=10000+1000+25=11025}$
3. ${\displaystyle 51^{2}-49^{2}=(51+49)(51-49)=100\times 2=200}$