啓林館：未来へ広がる数学３

p.13 問1\(\begin{split}{\small (1)}~14x^2+7xy\end{split}\)
\(\begin{split}{\small (2)}~12a^2-4ab\end{split}\)
\(\begin{split}{\small (3)}~-10ab+12b^2\end{split}\)
\(\begin{split}{\small (4)}~8x^2-4x\end{split}\)
\(\begin{split}{\small (5)}~2x^2+6xy\end{split}\)
\(\begin{split}{\small (6)}~-24a^2-21ab\end{split}\)
\(\begin{split}{\small (7)}~6x^2-4xy\end{split}\)
\(\begin{split}{\small (8)}~4x^2-12xy-8x\end{split}\)
\(\begin{split}{\small (9)}~-12x^2+9xy-6x\end{split}\)
\(\begin{split}{\small (10)}~-3a^2+6ab-3a\end{split}\)

p.13 問2\(\begin{split}{\small (1)}~x-2\end{split}\)　　\(\begin{split}{\small (2)}~4a-1\end{split}\)
\(\begin{split}{\small (3)}~-2x-y\end{split}\)　　\(\begin{split}{\small (4)}~3a-b\end{split}\)
\(\begin{split}{\small (5)}~15x-10xy\end{split}\)　　\(\begin{split}{\small (6)}~-20x+2\end{split}\)
\(\begin{split}{\small (7)}~-4x-8y\end{split}\)　　\(\begin{split}{\small (8)}~6x-4y\end{split}\)

p.14 問3\(\begin{split}{\small (1)}~ac-ad+bc-bd\end{split}\)
\(\begin{split}{\small (2)}~ac-ad-bc+bd\end{split}\)
\(\begin{split}{\small (3)}~xy+3x+2y+6\end{split}\)
\(\begin{split}{\small (4)}~xy+4x-y-4\end{split}\)

p.15 問4\(\begin{split}{\small (1)}~x^2-8x+12\end{split}\)
\(\begin{split}{\small (2)}~x^2+x-20\end{split}\)
\(\begin{split}{\small (3)}~a^2-2a-3\end{split}\)
\(\begin{split}{\small (4)}~a^2+15a+56\end{split}\)

p.15 問5\(\begin{split}{\small (1)}~6a^2+13ab+6b^2\end{split}\)
\(\begin{split}{\small (2)}~45a^2+44ab-12b^2\end{split}\)
\(\begin{split}{\small (3)}~7x^2-31xy-20y^2\end{split}\)
\(\begin{split}{\small (4)}~16x^2-26xy+3y^2\end{split}\)

p.15 問6\(\begin{split}{\small (1)}~a^2+ab+b-1\end{split}\)
\(\begin{split}{\small (2)}~2a^2+5ab+a+2b^2+2b\end{split}\)
\(\begin{split}{\small (3)}~2x^2+3xy-2x-2y^2+y\end{split}\)
\(\begin{split}{\small (4)}~3x^2-5xy+9x+2y^2-6y\end{split}\)

p.16 問1\(\begin{split}{\small (1)}~x^2+5x+6\end{split}\)
\(\begin{split}{\small (2)}~x^2-10x+24\end{split}\)
\(\begin{split}{\small (3)}~x^2+4x-45\end{split}\)
\(\begin{split}{\small (4)}~x^2-3x-40\end{split}\)

p.17 問2\(\begin{split}{\small (1)}~a^2+6a+9\end{split}\)
\(\begin{split}{\small (2)}~x^2-14x+49\end{split}\)
\(\begin{split}{\small (3)}~y^2+8y+16\end{split}\)

p.17 問3\(\begin{split}{\small (1)}~x^2-10xy+25y^2\end{split}\)
\(\begin{split}{\small (2)}~a^2+8ab+16b^2\end{split}\)
\(\begin{split}{\small (3)}~16x^2-8xy+y^2\end{split}\)
\(\begin{split}{\small (4)}~4x^2+12xy+9y^2\end{split}\)

---

\(\begin{split}{\small (5)}~a^2+ab+\frac{\,1\,}{\,4\,}b^2\end{split}\)

---

\(\begin{split}{\small (6)}~x^2-4xy+4y^2\end{split}\)

p.18 問4\(\begin{split}{\small (1)}~x^2-64\end{split}\)
\(\begin{split}{\small (2)}~9-a^2\end{split}\)
\(\begin{split}{\small (3)}~25x^2-1\end{split}\)
\(\begin{split}{\small (4)}~9x^2-4y^2\end{split}\)

---

\(\begin{split}{\small (5)}~x^2-\frac{\,1\,}{\,9\,}\end{split}\)

---

\(\begin{split}{\small (6)}~a^2-36b^2\end{split}\)

p.18 問5[証明] 左のカードの数を \(x\) として、
① \(x+5=A\)
② \(x-5=B\)
③ \(5^2=25=C\)
④ \(A\times B+C=D\) より、
\(\begin{split}&(x+5)(x-5)+25
\\[2pt]~~=~&x^2-25+25
\\[2pt]~~=~&x^2
\end{split}\)
よって、\(D=x^2\) となるので、左の数の２乗となる [終]

p.19 問6\(\begin{split}{\small (1)}~2x^2+2\end{split}\)　　\(\begin{split}{\small (2)}~x+18\end{split}\)

p.19 問7\(\begin{split}{\small (1)}~a^2+2ab+b^2+2a+2b-3\end{split}\)
\(\begin{split}{\small (2)}~x^2+2xy+y^2-4x-4y+4\end{split}\)

p.20 練習問題1.\(\begin{split}{\small (1)}~x^2+11x+28\end{split}\)
　\(\begin{split}{\small (2)}~x^2-7x-8\end{split}\)
　\(\begin{split}{\small (3)}~x^2-13xy+36y^2\end{split}\)
　\(\begin{split}{\small (4)}~x^2+8x+16\end{split}\)
　\(\begin{split}{\small (5)}~9x^2-12x+4\end{split}\)
　\(\begin{split}{\small (6)}~16x^2-24x+9y^2\end{split}\)

---

　\(\begin{split}{\small (7)}~\frac{\,1\,}{\,4\,}x^2+2x+4\end{split}\)

---

　\(\begin{split}{\small (8)}~25-a^2\end{split}\)
　\(\begin{split}{\small (9)}~x^2-49y^2\end{split}\)
　\(\begin{split}{\small (10)}~1-2x+x^2\end{split}\)
　\(\begin{split}{\small (11)}~-25x^2+10x-1\end{split}\)

---

　\(\begin{split}{\small (12)}~a^2-\frac{\,1\,}{\,4\,}a-\frac{\,1\,}{\,8\,}\end{split}\)

---

2.\(\begin{split}{\small (1)}~12x-85\end{split}\)
　\(\begin{split}{\small (2)}~4ab\end{split}\)
　\(\begin{split}{\small (3)}~3x^2+4xy+10y^2\end{split}\)
　\(\begin{split}{\small (4)}~2x^2+13\end{split}\)

---

3.\(\begin{split}{\small (1)}~a^2-4ab+4b^2-9\end{split}\)
　\(\begin{split}{\small (2)}~x^2+2xy+y^2-14x-14y+49\end{split}\)
　\(\begin{split}{\small (3)}~x^2+2xy+y^2-5x-5y\end{split}\)
　\(\begin{split}{\small (4)}~x^2+6xy+9y^2-11x-33y+1\end{split}\)

p.22 問1\(\begin{split}{\small (1)}~a(b-c)\end{split}\)　　\(\begin{split}{\small (2)}~2a(2x-1)\end{split}\)
\(\begin{split}{\small (3)}~a(2x+3y)\end{split}\)　　\(\begin{split}{\small (4)}~4b(2a^2-b)\end{split}\)
\(\begin{split}{\small (5)}~ab(a-b)\end{split}\)　　\(\begin{split}{\small (6)}~x(a+b+c)\end{split}\)

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

p.23 問3\(\begin{split}{\small (1)}~(x+1)^2\end{split}\)　　\(\begin{split}{\small (2)}~(x-2)^2\end{split}\)
\(\begin{split}{\small (3)}~(x+7)^2\end{split}\)　　\(\begin{split}{\small (4)}~(x-6)^2\end{split}\)

p.23 問4\(\begin{split}{\small (1)}~(2x-3)^2\end{split}\)　　\(\begin{split}{\small (2)}~(4y+5)^2\end{split}\)
\(\begin{split}{\small (3)}~(3a-b)^2\end{split}\)　　\(\begin{split}{\small (4)}~(2t-5)^2\end{split}\)

p.23 問5\(\begin{split}{\small (1)}~x^2-6x+9=(x-3)^2\end{split}\)
\(\begin{split}{\small (2)}~4x^2+4x+1=(2x+1)^2\end{split}\)
\(\begin{split}{\small (3)}~x^2-16x+64=(x-8)^2\end{split}\)

p.24 問6\(\begin{split}{\small (1)}~(x+1)(x+2)\end{split}\)
\(\begin{split}{\small (2)}~(x+1)(x+6)\end{split}\)
\(\begin{split}{\small (3)}~(x+2)(x+6)\end{split}\)
\(\begin{split}{\small (4)}~(x+3)(x+8)\end{split}\)

p.24 問7\(\begin{split}{\small (1)}~(x-1)(x-3)\end{split}\)
\(\begin{split}{\small (2)}~(x-1)(x-7)\end{split}\)
\(\begin{split}{\small (3)}~(x-3)(x-6)\end{split}\)
\(\begin{split}{\small (4)}~(x-2)(x-8)\end{split}\)

p.24 問8\(\begin{split}{\small (1)}~(x+8)(x-1)\end{split}\)
\(\begin{split}{\small (2)}~(x+3)(x-2)\end{split}\)
\(\begin{split}{\small (3)}~(x+5)(x-2)\end{split}\)
\(\begin{split}{\small (4)}~(x+7)(x-5)\end{split}\)
\(\begin{split}{\small (5)}~(x+1)(x-9)\end{split}\)
\(\begin{split}{\small (6)}~(x+1)(x-10)\end{split}\)

p.25 問9\(\begin{split}{\small (1)}~(x+6)(x-5)\end{split}\)
\(\begin{split}{\small (2)}~(x+2)(x+5)\end{split}\)
\(\begin{split}{\small (3)}~(a-1)(a-4)\end{split}\)
\(\begin{split}{\small (4)}~(a+5)(a-3)\end{split}\)
\(\begin{split}{\small (5)}~(y+1)(y-2)\end{split}\)
\(\begin{split}{\small (6)}~(t+3)(t+7)\end{split}\)

p.26 問10\(\begin{split}{\small (1)}~5(x+3)(x-3)\end{split}\)
\(\begin{split}{\small (2)}~3a(x+2)^2\end{split}\)
\(\begin{split}{\small (3)}~2b(x+2)(x-4)\end{split}\)
\(\begin{split}{\small (4)}~b(2a+x)(2a-x)\end{split}\)

p.26 問11\(\begin{split}{\small (1)}~(a+b)(x+y)\end{split}\)
\(\begin{split}{\small (2)}~(x-a)(y-b)\end{split}\)
\(\begin{split}{\small (3)}~(x+1)(x-2)\end{split}\)
\(\begin{split}{\small (4)}~(a+b+2)(a+b+3)\end{split}\)

p.27 話しあおう\(\begin{split}{\small (1)}~(b-4)(2a+1)\end{split}\)
\(\begin{split}{\small (2)}~(x+1)(y+1)\end{split}\)

p.27 練習問題1.\(\begin{split}{\small (1)}~m(x-y)\end{split}\)
　\(\begin{split}{\small (2)}~2b(a-2b)\end{split}\)
　\(\begin{split}{\small (3)}~a(xy+y+1)\end{split}\)
　\(\begin{split}{\small (4)}~-7a(2a+3b-1)\end{split}\)
　\(\begin{split}{\small (5)}~6ab(3a-2)\end{split}\)
　\(\begin{split}{\small (6)}~4b(ac+4a-2c)\end{split}\)

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2.\(\begin{split}{\small (1)}~(x+5)^2\end{split}\)
　\(\begin{split}{\small (2)}~(a-7)^2\end{split}\)
　\(\begin{split}{\small (3)}~(x+8)(x-8)\end{split}\)
　\(\begin{split}{\small (4)}~(5a+4b)(5a-4b)\end{split}\)
　\(\begin{split}{\small (5)}~(2a+1)(2a-1)\end{split}\)
　\(\begin{split}{\small (6)}~(y-6)^2\end{split}\)

---

3.\(\begin{split}{\small (1)}~(x+1)(x+3)\end{split}\)
　\(\begin{split}{\small (2)}~(x+2)(x-1)\end{split}\)
　\(\begin{split}{\small (3)}~(x+2)(x-3)\end{split}\)
　\(\begin{split}{\small (4)}~(x+3)(x-6)\end{split}\)
　\(\begin{split}{\small (5)}~(x+7)(x-2)\end{split}\)
　\(\begin{split}{\small (6)}~(x+2)(x-8)\end{split}\)

---

4.\(\begin{split}{\small (1)}~(x-2)(x-6)\end{split}\)
　\(\begin{split}{\small (2)}~(a+3)(a-1)\end{split}\)
　\(\begin{split}{\small (3)}~(6y+7)^2\end{split}\)
　\(\begin{split}{\small (4)}~(10-y)^2\end{split}\)
　\(\begin{split}{\small (5)}~(2-a)(14-a)\end{split}\)
　\(\begin{split}{\small (6)}~(x+1)(x-3)\end{split}\)

---

5.\(\begin{split}{\small (1)}~4(x+2)(x-5)\end{split}\)
　\(\begin{split}{\small (2)}~-3a(x-1)^2\end{split}\)
　\(\begin{split}{\small (3)}~y(x+1)(x-1)\end{split}\)
　\(\begin{split}{\small (4)}~(x+y)(a-3)\end{split}\)
　\(\begin{split}{\small (5)}~(a+b-2)^2\end{split}\)
　\(\begin{split}{\small (6)}~(a-b+c)(a-b-c)\end{split}\)

p.29 問1\(\begin{split}~~~2\times4+1=9=3^2\end{split}\)
\(\begin{split}~~~4\times6+1=25=5^2\end{split}\)
\(\begin{split}~~~6\times8+1=49=7^2\end{split}\)
\(\begin{split}~~~8\times10+1=81=9^2\end{split}\)
\(\begin{split}~~~10\times12+1=121=11^2\end{split}\)
\(\begin{split}~~~12\times14+1=169=13^2\end{split}\)
\(\begin{split}~~~14\times16+1=225=15^2\end{split}\)

p.30 問2連続した２つの偶数の積に１たした数は、その２つの偶数の間の整数(奇数)の２乗となる

p.30 問3\(\begin{split}{\small (1)}~800\end{split}\)　　\(\begin{split}{\small (2)}~5200\end{split}\)　　\(\begin{split}{\small (3)}~29600\end{split}\)

p.30 問4\(\begin{split}{\small (1)}~10404\end{split}\)　　\(\begin{split}{\small (2)}~1599\end{split}\)　　\(\begin{split}{\small (3)}~9801\end{split}\)

p.31 問5\(\begin{split}{\small (1)}~250\end{split}\)　　\(\begin{split}{\small (2)}~-100\end{split}\)

p.32 問6[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]