東京書籍：新しい数学３

p.13 問1\(\begin{split}{\small (1)}~4a^2-12ab\end{split}\)
\(\begin{split}{\small (2)}~-10x^2+35xy\end{split}\)
\(\begin{split}{\small (3)}~-5ab+b^2\end{split}\)
\(\begin{split}{\small (4)}~2a^2-2ab-2ac\end{split}\)
\(\begin{split}{\small (5)}~-18x^2-12xy+6x\end{split}\)
\(\begin{split}{\small (6)}~2x^2+4xy+6x\end{split}\)

p.13 問2\(\begin{split}{\small (1)}~5x^2+7x\end{split}\)　　\(\begin{split}{\small (2)}~-2a^2\end{split}\)

p.13 問3\(\begin{split}{\small (1)}~3a^2-1\end{split}\)　　\(\begin{split}{\small (2)}~-4a^2-1\end{split}\)
\(\begin{split}{\small (3)}~4a-6b\end{split}\)　　\(\begin{split}{\small (4)}~xy+y^2-1\end{split}\)

p.15 問1\(\begin{split}{\small (1)}~xy+2x+6y+12\end{split}\)
\(\begin{split}{\small (2)}~ab+2a-3b-6\end{split}\)
\(\begin{split}{\small (3)}~ac-ad-bc+bd\end{split}\)
\(\begin{split}{\small (4)}~2xy-14x+y-7\end{split}\)

p.15 問2\(\begin{split}{\small (1)}~x^2+6x+8\end{split}\)
\(\begin{split}{\small (2)}~x^2-5x+6\end{split}\)
\(\begin{split}{\small (3)}~x^2-25\end{split}\)
\(\begin{split}{\small (4)}~12x^2-5x-2\end{split}\)
\(\begin{split}{\small (5)}~2a^2+7ab+3b^2\end{split}\)

p.15 問3\(\begin{split}{\small (1)}~a^2-ab+3a-b+2\end{split}\)
\(\begin{split}{\small (2)}~10x^2-xy-3y^2-5x+3y\end{split}\)

p.17 問1\(\begin{split}{\small (1)}~x^2+9x+18\end{split}\)　　\(\begin{split}{\small (2)}~x^2-2x-3\end{split}\)

p.17 問2\(\begin{split}{\small (1)}~x^2+3x+2\end{split}\)
\(\begin{split}{\small (2)}~x^2+4x-12\end{split}\)
\(\begin{split}{\small (3)}~x^2-7x+12\end{split}\)
\(\begin{split}{\small (4)}~y^2+8y+15\end{split}\)
\(\begin{split}{\small (5)}~a^2-15a+56\end{split}\)
\(\begin{split}{\small (6)}~x^2-x-30\end{split}\)
\(\begin{split}{\small (7)}~x^2+0.2x-0.08\end{split}\)

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\(\begin{split}{\small (8)}~y^2-\frac{\,1\,}{\,3\,}y-\frac{\,2\,}{\,9\,}\end{split}\)

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p.18 問3\(\begin{split}{\small (1)}~x^2+12x+36\end{split}\)
\(\begin{split}{\small (2)}~a^2+18a+81\end{split}\)
\(\begin{split}{\small (3)}~x^2-10x+25\end{split}\)
\(\begin{split}{\small (4)}~y^2-14y+49\end{split}\)

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\(\begin{split}{\small (5)}~x^2+\frac{\,2\,}{\,3\,}x+\frac{\,1\,}{\,9\,}\end{split}\)

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

p.19 問4\(\begin{split}{\small (1)}~x^2-9\end{split}\)　　\(\begin{split}{\small (2)}~x^2-25\end{split}\)

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\(\begin{split}{\small (3)}~x^2-\frac{\,1\,}{\,9\,}\end{split}\)　　\(\begin{split}{\small (4)}~4-x^2\end{split}\)

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\(\begin{split}{\small (5)}~a^2-b^2\end{split}\)

p.19 問5\(\begin{split}{\small (1)}~x^2-8x+16\end{split}\)
\(\begin{split}{\small (2)}~x^2-2x-24\end{split}\)
\(\begin{split}{\small (3)}~x^2-49\end{split}\)
\(\begin{split}{\small (4)}~a^2+2ab+b^2\end{split}\)
\(\begin{split}{\small (5)}~x^2+8x+12\end{split}\)
\(\begin{split}{\small (6)}~64+16a+a^2\end{split}\)
\(\begin{split}{\small (7)}~a^2+3a-10\end{split}\)
\(\begin{split}{\small (8)}~x^2-81\end{split}\)

p.20 問6\(\begin{split}{\small (1)}~9x^2-18x+8\end{split}\)
\(\begin{split}{\small (2)}~16a^2+12a-18\end{split}\)

p.20 問7\(\begin{split}{\small (1)}~25x^2+20x+4\end{split}\)
\(\begin{split}{\small (2)}~9x^2-24xy+16y^2\end{split}\)
\(\begin{split}{\small (3)}~36x^2-49\end{split}\)
\(\begin{split}{\small (4)}~49a^2-16b^2\end{split}\)

p.20 問8２行目で \(2\times5\times x\) と \(x\) を使っているが、正しくは \(3x\) を使う
\(\begin{split}&(3x+5)^2
\\[2pt]~~=~&(3x)^2+2\times 5\times (3x)+5^2
\\[2pt]~~=~&9x^2+30x+25
\end{split}\)

p.21 問9\(\begin{split}{\small (1)}~x^2+2xy+y^2-2x-2y-15\end{split}\)
\(\begin{split}{\small (2)}~a^2+2ab+b^2+2ac+2bc+c^2\end{split}\)
\(\begin{split}{\small (3)}~a^2-2ab+b^2-12a+12b+36\end{split}\)
\(\begin{split}{\small (4)}~a^2+6a+9-b^2\end{split}\)

p.21 問10\(\begin{split}{\small (1)}~2x^2+x+8\end{split}\)　　\(\begin{split}{\small (2)}~x^2+x+4\end{split}\)

p.25 問1\(\begin{split}{\small (1)}~x(a-b)\end{split}\)　　\(\begin{split}{\small (2)}~b(2b+a)\end{split}\)
\(\begin{split}{\small (3)}~a(x^2+x+2)\end{split}\)

p.25 問2\(\begin{split}{\small (1)}~2x(3m-n)\end{split}\)　　\(\begin{split}{\small (2)}~5x(2xy+1)\end{split}\)
\(\begin{split}{\small (3)}~xy(y-x)\end{split}\)　　\(\begin{split}{\small (4)}~2ab(2a-3b-5)\end{split}\)

p.26 問1\(\begin{split}{\small (1)}~(x+2)(x+5)\end{split}\)
\(\begin{split}{\small (2)}~(x-1)(x-6)\end{split}\)
\(\begin{split}{\small (3)}~(x+2)(x+6)\end{split}\)
\(\begin{split}{\small (4)}~(x-1)(x-8)\end{split}\)

p.27 問2\(\begin{split}{\small (1)}~(x+3)(x-5)\end{split}\)
\(\begin{split}{\small (2)}~(x+2)(x-4)\end{split}\)
\(\begin{split}{\small (3)}~(x+5)(x-2)\end{split}\)
\(\begin{split}{\small (4)}~(a-8)(a+1)\end{split}\)

p.27 問3\(\begin{split}{\small (1)}~(x+1)(x+9)\end{split}\)
\(\begin{split}{\small (2)}~(y+9)(y-4)\end{split}\)
\(\begin{split}{\small (3)}~(x+4)(x-7)\end{split}\)
\(\begin{split}{\small (4)}~(x-2)(x-14)\end{split}\)
\(\begin{split}{\small (5)}~(x+2)(x-1)\end{split}\)
\(\begin{split}{\small (6)}~(x+1)(x+100)\end{split}\)

p.27 問4\(\begin{split}{\small (1)}~(x+6)^2\end{split}\)　　\(\begin{split}{\small (2)}~(x+2)^2\end{split}\)
\(\begin{split}{\small (3)}~(a+9)^2\end{split}\)　　\(\begin{split}{\small (4)}~(x-1)^2\end{split}\)
\(\begin{split}{\small (5)}~(y-7)^2\end{split}\)　　\(\begin{split}{\small (6)}~(x-8)^2\end{split}\)

p.28 問5\(\begin{split}{\small (1)}~(x+6)(x-6)\end{split}\)
\(\begin{split}{\small (2)}~(a+2)(a-2)\end{split}\)
\(\begin{split}{\small (3)}~(x+1)(x-1)\end{split}\)
\(\begin{split}{\small (4)}~(4+y)(4-y)\end{split}\)

p.28 問6\(\begin{split}{\small (1)}~(x-1)(x-2)\end{split}\)
\(\begin{split}{\small (2)}~(x+8)(x-8)\end{split}\)
\(\begin{split}{\small (3)}~(y-2)^2\end{split}\)
\(\begin{split}{\small (4)}~(x+3)(x-8)\end{split}\)
\(\begin{split}{\small (5)}~(x+4)(x+9)\end{split}\)
\(\begin{split}{\small (6)}~(a+11)^2\end{split}\)

p.28 問7\(\begin{split}{\small (1)}~\end{split}\)\(x\) の１次式がないので、
\(\begin{split}~~~~x^2-a^2=(x+a)(x-a)\end{split}\)
この公式で \(a=4\) として、
\(\begin{split}&x^2-16
\\[2pt]~~=~&(x+4)(x-4)
\end{split}\)

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\(\begin{split}{\small (2)}~\end{split}\)\(8=2\times 4~,~16=4^2\) で \(x\) の１次式が正の項であるので、
\(\begin{split}~~~~x^2+2ax+a^2=(x+a)^2\end{split}\)
この公式で \(a=4\) として、
\(\begin{split}&x^2+8x+16
\\[2pt]~~=~&(x+4)^2
\end{split}\)

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\(\begin{split}{\small (3)}~\end{split}\)\(8=2\times 4~,~16=4^2\) で \(x\) の１次式が負の項であるので、
\(\begin{split}~~~~x^2-2ax+a^2=(x-a)^2\end{split}\)
この公式で \(a=4\) として、
\(\begin{split}&x^2-8x+16
\\[2pt]~~=~&(x-4)^2
\end{split}\)

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\(\begin{split}{\small (4)}~\end{split}\)和が \(-10\)、積が \(16\) となる２つの数は \(-2\) と \(-8\) である
\(\begin{split}&x^2+(a+b)x+ab
\\[2pt]~~=~&(x+a)(x+b)
\end{split}\)
この公式で \(a=-2~,~b=-8\) として、
\(\begin{split}&x^2-10x+16
\\[2pt]~~=~&(x-2)(x-8)
\end{split}\)

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

p.29 問9\(\begin{split}{\small (1)}~(3x+1)^2\end{split}\)
\(\begin{split}{\small (2)}~(x-10y)^2\end{split}\)
\(\begin{split}{\small (3)}~(x+7y)(x-7y)\end{split}\)
\(\begin{split}{\small (4)}~(2a+5b)(2a-5b)\end{split}\)

p.29 問10\(\begin{split}{\small (1)}~2y(x-1)(x-3)\end{split}\)
\(\begin{split}{\small (2)}~4(x+3y)(x-3y)\end{split}\)

p.29 問11\(\begin{split}{\small (1)}~(a-2)(x+y)\end{split}\)
\(\begin{split}{\small (2)}~(a+b+2)(a+b+3)\end{split}\)
\(\begin{split}{\small (3)}~(a-1)(a-8)\end{split}\)
\(\begin{split}{\small (4)}~(3x+4)(x+10)\end{split}\)

p.33 問1\(\begin{split}{\small (1)}~9604\end{split}\)　　\(\begin{split}{\small (2)}~3600\end{split}\)　　\(\begin{split}{\small (3)}~2491\end{split}\)

p.33 問2\(\begin{split}~~~1600\end{split}\)