数研出版：これからの数学３

p.14 問1\(\begin{split}{\small (1)}~2ab+ac\end{split}\)
\(\begin{split}{\small (2)}~-2x^2+2x\end{split}\)
\(\begin{split}{\small (3)}~15ac-20bc\end{split}\)
\(\begin{split}{\small (4)}~4x^2+12xy-8y\end{split}\)
\(\begin{split}{\small (5)}~6ab-12b^2-18b\end{split}\)
\(\begin{split}{\small (6)}~-2x^3+4x^2-x\end{split}\)

p.15 問2\(\begin{split}{\small (1)}~2a-3\end{split}\)　　\(\begin{split}{\small (2)}~4x-1\end{split}\)
\(\begin{split}{\small (3)}~-9a-21b\end{split}\)　　\(\begin{split}{\small (4)}~8x-12y\end{split}\)

p.15 問3\(\begin{split}{\small (1)}~-a^2+3a\end{split}\)　　\(\begin{split}{\small (2)}~x^2-4\end{split}\)

p.16 問1\(a+b=N\) とすると、
\(\begin{split}&(a+b)(c+d)
\\[2pt]~~=~&N(c+d)
\\[2pt]~~=~&cN+dN
\\[2pt]~~=~&c(a+b)+d(a+b)
\\[2pt]~~=~&ac+bc+ad+bd
\end{split}\)

p.17 問2\(\begin{split}{\small (1)}~xy+5x+3y+15\end{split}\)
\(\begin{split}{\small (2)}~ab-7a+2b-14\end{split}\)
\(\begin{split}{\small (3)}~ax-ay-bx+by\end{split}\)

p.17 問3\(\begin{split}{\small (1)}~x^2+3x-4\end{split}\)
\(\begin{split}{\small (2)}~2a^2+7a+3\end{split}\)
\(\begin{split}{\small (3)}~-x^2+x+6\end{split}\)
\(\begin{split}{\small (4)}~-15a^2+11ab-2b^2\end{split}\)

p.17 問4\(\begin{split}{\small (1)}~a^2+3ab-a-6b-2\end{split}\)
\(\begin{split}{\small (2)}~8x^2-2xy-3y^2+2x+y\end{split}\)

p.19 問1\(\begin{split}{\small (1)}~x^2+7x+12\end{split}\)
\(\begin{split}{\small (2)}~x^2+4x-5\end{split}\)
\(\begin{split}{\small (3)}~x^2-x-6\end{split}\)
\(\begin{split}{\small (4)}~x^2-10x+24\end{split}\)
\(\begin{split}{\small (5)}~y^2+9y+14\end{split}\)
\(\begin{split}{\small (6)}~t^2-14t+48\end{split}\)
\(\begin{split}{\small (7)}~a^2+7a-18\end{split}\)

---

\(\begin{split}{\small (8)}~x^2+x+\frac{\,2\,}{\,9\,}\end{split}\)

---

\(\begin{split}{\small (9)}~x^2-\frac{\,2\,}{\,5\,}x-\frac{\,3\,}{\,25\,}\end{split}\)

---

\(\begin{split}{\small (10)}~-63-2x+x^2\end{split}\)

p.20 問2\(\begin{split}{\small (1)}~a^2+2a+1\end{split}\)
\(\begin{split}{\small (2)}~x^2+8x+16\end{split}\)
\(\begin{split}{\small (3)}~x^2-6x+9\end{split}\)
\(\begin{split}{\small (4)}~y^2-14y+49\end{split}\)

---

\(\begin{split}{\small (5)}~x^2+\frac{\,4\,}{\,5\,}x+\frac{\,4\,}{\,25\,}\end{split}\)

---

\(\begin{split}{\small (6)}~x^2-2xy+y^2\end{split}\)

p.21 問3\(\begin{split}{\small (1)}~4x^2+4x+1\end{split}\)
\(\begin{split}{\small (2)}~x^2-6xy+9y^2\end{split}\)
\(\begin{split}{\small (3)}~9x^2-30xy+25y^2\end{split}\)

p.21 問4\(\begin{split}{\small (1)}~a^2-36\end{split}\)　　\(\begin{split}{\small (2)}~x^2-1\end{split}\)

---

\(\begin{split}{\small (3)}~x^2-49\end{split}\)　　\(\begin{split}{\small (4)}~x^2-\frac{\,9\,}{\,25\,}\end{split}\)

---

\(\begin{split}{\small (5)}~m^2-25\end{split}\)　　\(\begin{split}{\small (6)}~9-a^2\end{split}\)

p.21 問5\(\begin{split}{\small (1)}~16x^2-y^2\end{split}\)　　\(\begin{split}{\small (2)}~9x^2-4y^2\end{split}\)
\(\begin{split}{\small (3)}~49-25x^2\end{split}\)

p.22 問6\(\begin{split}{\small (1)}~4x^2-16x+7\end{split}\)
\(\begin{split}{\small (2)}~9x^2-9xy+2y^2\end{split}\)
\(\begin{split}{\small (3)}~x^2+2xy-3y^2\end{split}\)

p.23 問7\(\begin{split}{\small (1)}~a^2-2ab+b^2+3a-3b-4\end{split}\)
\(\begin{split}{\small (2)}~x^2+2xy+y^2+2x+2y+1\end{split}\)
\(\begin{split}{\small (3)}~x^2+4xy+4y^2-9\end{split}\)

p.23 問8\(\begin{split}{\small (1)}~3x^2-2x+9\end{split}\)
\(\begin{split}{\small (2)}~-2x-2\end{split}\)
\(\begin{split}{\small (3)}~8a^2-4a+1\end{split}\)

p.27 問1\(\begin{split}{\small (1)}~b(a+c)\end{split}\)
\(\begin{split}{\small (2)}~xy(x-2)\end{split}\)
\(\begin{split}{\small (3)}~6a(ax+2y)\end{split}\)
\(\begin{split}{\small (4)}~m(3x+4y-2z)\end{split}\)
\(\begin{split}{\small (5)}~2ab(4b-a+2)\end{split}\)

p.29 問1\(\begin{split}{\small (1)}~(x+1)(x+3)\end{split}\)
\(\begin{split}{\small (2)}~(x+3)(x+5)\end{split}\)
\(\begin{split}{\small (3)}~(x+3)(x+4)\end{split}\)
\(\begin{split}{\small (4)}~(x-1)(x-12)\end{split}\)
\(\begin{split}{\small (5)}~(x-2)(x-6)\end{split}\)
\(\begin{split}{\small (6)}~(x-2)(x-8)\end{split}\)

p.29 問2\(\begin{split}{\small (1)}~(x+1)(x-2)\end{split}\)
\(\begin{split}{\small (2)}~(x+5)(x-1)\end{split}\)
\(\begin{split}{\small (3)}~(x+1)(x-6)\end{split}\)
\(\begin{split}{\small (4)}~(x+4)(x-3)\end{split}\)
\(\begin{split}{\small (5)}~(y+3)(y-7)\end{split}\)
\(\begin{split}{\small (6)}~(a+9)(a-2)\end{split}\)

p.30 問3\(\begin{split}{\small (1)}~(x+5)^2\end{split}\)　　\(\begin{split}{\small (2)}~(x-4)^2\end{split}\)
\(\begin{split}{\small (3)}~(y+2)^2\end{split}\)　　\(\begin{split}{\small (4)}~(a-11)^2\end{split}\)

p.31 問4\(\begin{split}{\small (1)}~(3x+1)^2\end{split}\)
\(\begin{split}{\small (2)}~(2x-5)^2\end{split}\)
\(\begin{split}{\small (3)}~(x+2y)^2\end{split}\)
\(\begin{split}{\small (4)}~(3x-2y)^2\end{split}\)

p.31 問5\(\begin{split}{\small (1)}~(x+3)(x-3)\end{split}\)
\(\begin{split}{\small (2)}~(x+1)(x-1)\end{split}\)
\(\begin{split}{\small (3)}~(y+2)(y-2)\end{split}\)
\(\begin{split}{\small (4)}~(a+6)(a-6)\end{split}\)
\(\begin{split}{\small (5)}~(4+x)(4-x)\end{split}\)
\(\begin{split}{\small (6)}~(7+a)(7-a)\end{split}\)

p.32 問6\(\begin{split}{\small (1)}~(x+2y)(x-2y)\end{split}\)
\(\begin{split}{\small (2)}~(5a+6b)(5a-6b)\end{split}\)

p.33 問7\(\begin{split}{\small (1)}~3(x+2)^2\end{split}\)
\(\begin{split}{\small (2)}~a(b+1)(b-1)\end{split}\)
\(\begin{split}{\small (3)}~2y(x-2)(x-7)\end{split}\)
\(\begin{split}{\small (4)}~-2a(b-1)(b-2)\end{split}\)

p.34 問8\(\begin{split}{\small (1)}~(a+b)(a+b-5)\end{split}\)
\(\begin{split}{\small (2)}~(x+y+2)(x+y-4)\end{split}\)
\(\begin{split}{\small (3)}~x(x+2)\end{split}\)
\(\begin{split}{\small (4)}~(x-y+2)(x-y-2)\end{split}\)

p.35 問1\(\begin{split}{\small (1)}~2499\end{split}\)　　\(\begin{split}{\small (2)}~120\end{split}\)　　\(\begin{split}{\small (3)}~9409\end{split}\)

p.36 問2\(\begin{split}~~~2500\end{split}\)

p.36 問3\(\begin{split}~~~50\end{split}\)

p.37 問4連続する２つの奇数は、整数 \(n\) を使って、
\(\begin{split}~~~~~~2n-1~,~2n+1\end{split}\)
と表される
このとき、これらの積に \(1\) を加えたものは、
\(\begin{split}&(2n-1)(2n+1)+1
\\[2pt]~~=~&4n^2-1+1
\\[2pt]~~=~&4n^2
\\[2pt]~~=~&(2n)^2
\end{split}\)
\(2n\) は偶数であるから、連続する２つの奇数の積に \(1\) を加えると偶数の２乗となる

p.39 問5[証明] 道と花だん全体の面積は、一辺が \(p+2a\) より、
\(\begin{split}&(p+2a)(p+2a)
\\[2pt]~~=~&p^2+4ap+4a^2
\end{split}\)
また、花だんの面積は、
\(\begin{split}~~~~~~p\times p=p^2\end{split}\)
よって、道の面積 \(S\) は、
\(\begin{eqnarray}~S&=&(p^2+4ap+4a^2)-p^2
\\[2pt]~~~&=&4ap+4a^2~~~\cdots{\Large ①}
\end{eqnarray}\)
次に、道の一辺の長さは、
\(\begin{split}~~~~~~p+\frac{\,1\,}{\,2\,}a+\frac{\,1\,}{\,2\,}a=p+a\end{split}\)
よって、道のまん中を通る線の長さ \(l\) は、
\(\begin{eqnarray}~l&=&(p+a)\times 4
\\[2pt]~~~&=&4p+4a
\end{eqnarray}\)
これより、
\(\begin{eqnarray}~al&=&a(4p+4a)
\\[2pt]~~~&=&4ap+4a^2~~~\cdots{\Large ②}
\end{eqnarray}\)
したがって、①と②より、
\(\begin{split}~~~~~~S=al\end{split}\) [終]