{
  "api_version": "v1",
  "publisher": {
    "name": "Manazoo",
    "url": "https://manazoo.com/"
  },
  "license": "All rights reserved（AI の参照・要約と出典リンクつきの引用は可。転載・再配布・販売は不可）",
  "terms_url": "https://manazoo.com/terms/",
  "disclaimer": "内容の正確さには気をつけていますが、すべてが正しいとは限りません。テストや受験の前には、教科書や先生の説明でも確かめてください。間違いを見つけたら意見箱から知らせてください。",
  "attribution": "事実データの一部は Wikidata（CC0）による。単元の構成は文部科学省「中学校学習指導要領」「高等学校学習指導要領」（政府標準利用規約2.0）。",
  "kind": "quiz",
  "unit": {
    "slug": "various-expressions",
    "title": "いろいろな式",
    "url": "https://manazoo.com/high/math-2/various-expressions/"
  },
  "canonical_url": "https://manazoo.com/high/math-2/various-expressions/quiz/",
  "questions": [
    {
      "id": "q-bt-coeff-a2b2",
      "type": "input",
      "prompt": "(a＋b)⁴の展開式で、a²b²の項の係数を求めよ。",
      "choices": null,
      "answer": "6",
      "accept": [],
      "explanation": "a²b²の項はn＝4、r＝2の一般項₄C₂a²b²。₄C₂＝(4×3)/(2×1)＝6なので、係数は6。",
      "entities": [
        "binomial-theorem"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-bt-coeff-a3b2-minus",
      "type": "choice",
      "prompt": "(a−b)⁵の展開式で、a³b²の項の係数はどれか。",
      "choices": [
        "−10",
        "20",
        "10",
        "5"
      ],
      "answer": 2,
      "accept": [],
      "explanation": "(a−b)⁵＝(a＋(−b))⁵と考えると、a³b²の項は₅C₂a³(−b)²＝₅C₂a³b²。(−b)²はbの2乗なのでプラスになり、₅C₂＝(5×4)/(2×1)＝10。",
      "entities": [
        "binomial-theorem"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-bt-coeff-x4",
      "type": "input",
      "prompt": "(x＋1)⁶の展開式で、x⁴の項の係数を求めよ。",
      "choices": null,
      "answer": "15",
      "accept": [],
      "explanation": "一般項は₆Cᵣx⁶⁻ʳ1ʳ。x⁴の項は6−r＝4よりr＝2なので、係数は₆C₂＝(6×5)/(2×1)＝15。",
      "entities": [
        "binomial-theorem"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-bt-ncr-meaning",
      "type": "choice",
      "prompt": "二項定理の係数に出てくるₙCᵣが表しているものはどれか。",
      "choices": [
        "n個からr個を選ぶ選び方の数",
        "nをrで割った余り",
        "n個を一列に並べる並べ方の数",
        "n個のうちr個が同じものの個数"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "ₙCᵣは、n個のものの中からr個を選ぶ選び方の数を表す。並べ方の数（順列）や割り算の余りとは別のものなので、2つ目・3つ目は誤り。",
      "entities": [
        "binomial-theorem"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cef-a-minus-1-cubed",
      "type": "input",
      "prompt": "a＝4のとき、(a−1)³の値を求めよ。",
      "choices": null,
      "answer": "27",
      "accept": [],
      "explanation": "(a−1)³＝a³−3a²＋3a−1に代入すると64−48＋12−1＝27。(a−1)＝3を直接3乗しても27になる。",
      "entities": [
        "cubic-expansion-and-factoring"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cef-factor-x3-minus-64",
      "type": "choice",
      "prompt": "x³−64を因数分解すると、どれになるか。",
      "choices": [
        "(x＋4)(x²−4x＋16)",
        "(x−4)(x²＋4x−16)",
        "(x−4)(x²＋4x＋16)",
        "(x−4)(x²−4x＋16)"
      ],
      "answer": 2,
      "accept": [],
      "explanation": "64＝4³なので、a＝x、b＝4としてa³−b³＝(a−b)(a²＋ab＋b²)に当てはめると(x−4)(x²＋4x＋16)になる。2つ目はab項の符号を取り違えた式、3つ目はx³＋64、4つ目はx³−32x＋64になり、どれもx³−64と一致しない。",
      "entities": [
        "cubic-expansion-and-factoring"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cef-factor-x3-plus-1",
      "type": "choice",
      "prompt": "x³＋1を因数分解すると、どれになるか。",
      "choices": [
        "(x＋1)(x²−x＋1)",
        "(x−1)(x²＋x＋1)",
        "(x＋1)(x²＋x＋1)",
        "(x＋1)(x²−x−1)"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "1＝1³なので、a＝x、b＝1としてa³＋b³＝(a＋b)(a²−ab＋b²)に当てはめると(x＋1)(x²−x＋1)になる。2つ目はx³＋2x²＋2x＋1、3つ目はx³−1、4つ目はx³−2x−1になり、どれもx³＋1と一致しない。",
      "entities": [
        "cubic-expansion-and-factoring"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cef-sum-of-cubes-value",
      "type": "input",
      "prompt": "a＝5、b＝2のとき、a³＋b³の値を求めよ。",
      "choices": null,
      "answer": "133",
      "accept": [],
      "explanation": "a³＋b³＝(a＋b)(a²−ab＋b²)に代入すると(5＋2)(25−10＋4)＝7×19＝133。直接計算しても5³＋2³＝125＋8＝133になる。",
      "entities": [
        "cubic-expansion-and-factoring"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cn-add-complex",
      "type": "choice",
      "prompt": "(2＋3i)＋(1−5i)を計算せよ。",
      "choices": [
        "1−2i",
        "3−2i",
        "3＋2i",
        "3＋8i"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "実部どうし2＋1＝3、虚部どうし3−5＝−2を足すので、3−2iになる。",
      "entities": [
        "complex-numbers"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cn-multiply-complex",
      "type": "choice",
      "prompt": "(1＋2i)(3−i)を計算せよ。",
      "choices": [
        "1＋5i",
        "3＋5i",
        "5＋5i",
        "5−5i"
      ],
      "answer": 2,
      "accept": [],
      "explanation": "展開すると3−i＋6i−2i²。i²＝−1だから−2i²＝2となり、3＋5i＋2＝5＋5iになる。",
      "entities": [
        "complex-numbers"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cn-real-imag-product",
      "type": "input",
      "prompt": "複素数3＋4iの実部と虚部の積を求めよ。",
      "choices": null,
      "answer": "12",
      "accept": [],
      "explanation": "実部は3、虚部は4なので、積は3×4＝12。",
      "entities": [
        "complex-numbers"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-cn-sqrt-negative",
      "type": "choice",
      "prompt": "√−18をiを使って表せ。",
      "choices": [
        "−3√2i",
        "3√2i",
        "3√2",
        "9i"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "−18＝18×(−1)だから、√−18＝√18 i。√18＝3√2なので、3√2iになる。",
      "entities": [
        "complex-numbers"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eip-amgm-minimum",
      "type": "choice",
      "prompt": "a＞0のとき、a＋4／aの最小値はどれか。",
      "choices": [
        "2",
        "4",
        "8",
        "16"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "相加相乗平均の関係より\\(a+\\frac{4}{a} \\geq 2\\sqrt{a \\cdot \\frac{4}{a}}=2\\sqrt{4}=4\\)。等号はa＝4／aよりa＝2のときに成り立つので、最小値は4。",
      "entities": [
        "equality-and-inequality-proofs"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eip-arithmetic-mean",
      "type": "input",
      "prompt": "a＝3、b＝12のとき、相加平均(a＋b)／2の値を求めよ。",
      "choices": null,
      "answer": "15/2",
      "accept": [
        "7.5"
      ],
      "explanation": "相加平均は(a＋b)／2に代入して(3＋12)／2＝15／2（＝7.5）。",
      "entities": [
        "equality-and-inequality-proofs"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eip-geometric-mean",
      "type": "input",
      "prompt": "a＝4、b＝9のとき、相乗平均√(ab)の値を求めよ。",
      "choices": null,
      "answer": "6",
      "accept": [],
      "explanation": "ab＝4×9＝36なので、相乗平均は√36＝6。",
      "entities": [
        "equality-and-inequality-proofs"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-eip-undetermined-coeff",
      "type": "choice",
      "prompt": "等式(x−a)(x＋3)＝x²＋bx−6がxについての恒等式であるとき、a、bの値はどれか。",
      "choices": [
        "a＝3、b＝1",
        "a＝2、b＝1",
        "a＝2、b＝−1",
        "a＝−2、b＝1"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "左辺を展開するとx²＋(3−a)x−3a。右辺x²＋bx−6と係数を比べ、−3a＝−6よりa＝2、b＝3−a＝1。",
      "entities": [
        "equality-and-inequality-proofs"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-pdf-fraction-simplify-choice",
      "type": "choice",
      "prompt": "分数式 (x²＋2x)／x を約分した結果はどれか（x≠0）。",
      "choices": [
        "x＋2",
        "x",
        "x²＋2",
        "2"
      ],
      "answer": 0,
      "accept": [],
      "explanation": "x²＋2x＝x(x＋2)なので、(x²＋2x)／x＝x＋2（x≠0）に約分できる。分子の一部だけを消す2つ目・3つ目・4つ目は誤り。",
      "entities": [
        "polynomial-division-and-fractions"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-pdf-fraction-value-at-5",
      "type": "input",
      "prompt": "分数式 (x²−9)／(x−3) を約分してから、x＝5のときの値を求めよ。",
      "choices": null,
      "answer": "8",
      "accept": [],
      "explanation": "x²−9＝(x＋3)(x−3)なので、(x²−9)／(x−3)＝x＋3（x≠3）に約分できる。x＝5のとき5＋3＝8。",
      "entities": [
        "polynomial-division-and-fractions"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-pdf-quotient-x2-7x10",
      "type": "choice",
      "prompt": "x²＋7x＋10をx＋2で割ったときの商を求めよ。",
      "choices": [
        "x＋2",
        "x＋7",
        "x−5",
        "x＋5"
      ],
      "answer": 3,
      "accept": [],
      "explanation": "x²＋7x＋10は(x＋2)(x＋5)に等しいので、x＋2で割ると商はx＋5、余りは0になる。",
      "entities": [
        "polynomial-division-and-fractions"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-pdf-remainder-x3-2x5",
      "type": "input",
      "prompt": "x³＋2x＋5をx＋1で割ったときの余りを求めよ。",
      "choices": null,
      "answer": "2",
      "accept": [],
      "explanation": "最高次からそろえて引き算をくり返すと、x³＋2x＋5＝(x＋1)(x²−x＋3)＋2となり、余りは2。",
      "entities": [
        "polynomial-division-and-fractions"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-qdr-discriminant-sign",
      "type": "choice",
      "prompt": "二次方程式x²−3x＋5＝0の判別式の符号から、解の種類を答えよ。",
      "choices": [
        "異なる2つの実数解",
        "解はない",
        "異なる2つの虚数解",
        "重解"
      ],
      "answer": 2,
      "accept": [],
      "explanation": "D＝(−3)²−4×1×5＝9−20＝−11＜0なので、異なる2つの虚数解をもつ。",
      "entities": [
        "quadratic-discriminant-and-root-relations"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-qdr-equation-from-roots",
      "type": "choice",
      "prompt": "3と−2を解とする、x²の係数が1の二次方程式を1つ求めよ。",
      "choices": [
        "x²−x＋6＝0",
        "x²＋x−6＝0",
        "x²−6x−1＝0",
        "x²−x−6＝0"
      ],
      "answer": 3,
      "accept": [],
      "explanation": "和は3＋(−2)＝1、積は3×(−2)＝−6。x²−(和)x＋(積)＝0より、x²−x−6＝0になる。",
      "entities": [
        "quadratic-discriminant-and-root-relations"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-qdr-sum-of-roots",
      "type": "input",
      "prompt": "二次方程式x²−7x＋10＝0の2つの解をα,βとするとき、α＋βの値を求めよ。",
      "choices": null,
      "answer": "7",
      "accept": [],
      "explanation": "α＋β＝−b/a＝−(−7)/1＝7。実際に解いても2＋5＝7で一致する。",
      "entities": [
        "quadratic-discriminant-and-root-relations"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-qdr-sum-of-squares",
      "type": "input",
      "prompt": "二次方程式x²−3x＋1＝0の2つの解をα,βとするとき、α²＋β²の値を求めよ。",
      "choices": null,
      "answer": "7",
      "accept": [],
      "explanation": "α＋β＝3、αβ＝1なので、α²＋β²＝(α＋β)²−2αβ＝9−2＝7。",
      "entities": [
        "quadratic-discriminant-and-root-relations"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-rft-box-volume",
      "type": "input",
      "prompt": "縦が高さより2cm短く、横が高さより3cm長い直方体で、容積が350cm³になる箱を作る。高さxcmの値を求めよ（x＞2）。",
      "choices": null,
      "answer": "7",
      "accept": [
        "7cm"
      ],
      "explanation": "x(x−2)(x＋3)＝350を整理すると、x³＋x²−6x−350＝0。x＝7を入れると343＋49−42−350＝0になり成り立つ。",
      "entities": [
        "remainder-and-factor-theorems"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-rft-cubic-equation-roots",
      "type": "choice",
      "prompt": "x³−2x²−5x＋6＝0の解をすべて求めよ。",
      "choices": [
        "x＝1,−3,2",
        "x＝1,3,−2",
        "x＝1,3,2",
        "x＝−1,3,−2"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "P(1)＝1−2−5＋6＝0なのでx−1で割れる。商はx²−x−6＝(x−3)(x＋2)。解はx＝1,3,−2。",
      "entities": [
        "remainder-and-factor-theorems"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-rft-factor-theorem-check",
      "type": "choice",
      "prompt": "整式P(x)＝x³−4x²＋x＋6で、x＋1はP(x)の因数か。P(1)やP(−1)の値から判断せよ。",
      "choices": [
        "P(−1)＝4なので因数でない",
        "P(−1)＝0なので因数である",
        "P(1)＝0なので因数である",
        "P(1)＝4なので因数でない"
      ],
      "answer": 1,
      "accept": [],
      "explanation": "P(−1)＝−1−4−1＋6＝0なので、因数定理よりx＋1はP(x)の因数である。",
      "entities": [
        "remainder-and-factor-theorems"
      ],
      "verified": true,
      "placeholder": false
    },
    {
      "id": "q-rft-remainder-value",
      "type": "input",
      "prompt": "整式P(x)＝x³＋3x−5をx−1で割った余りを求めよ。",
      "choices": null,
      "answer": "-1",
      "accept": [
        "−1",
        "－1"
      ],
      "explanation": "剰余の定理より、余りはP(1)＝1＋3−5＝−1。",
      "entities": [
        "remainder-and-factor-theorems"
      ],
      "verified": true,
      "placeholder": false
    }
  ]
}