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1b:Tb2c,こんにちは！方程式の解き方を説明していきますね。
まず、最も重要なのは問題の状況把握です。私は塾講師もやっているのですが、ここでつまづいてしまうことがほとんどです。ここで、毎回生徒には「図を使って問題文を表してください」と伝えています。愛ちゃんさんは問題文を図で表すのは得意ですかね？例えば、全体を一つの棒グラフで表し、その比率を棒グラフの中に書き込む、池の周りを走るような問題であれば、池を書いてその周りに小さい人を走らせる、など自分がわかりやすいように図を描いてみてください！そうすれば、式に直しやしやすくなりますよ。

次に注目する箇所は何を文字で置くかです。これは、基本的には問題で聞かれているものを文字で置けば大丈夫です！
例題を参考にしてみましょう。
(例題)
同じ値段のボールペンを20本買おうとしたところ、持っているお金では400円足りなかったので、15本買おうとしたが、それでも50円足りなかった。この時ボールペン1本の値段を求めなさい。

①何を文字で置くか考える。
この問題では、ボールペン1本の値段を聞かれているので、これをXと置くことにする。

②状況把握する。
ここで、何円自分が持っているのかわからないことに気づく。そのため、分からないものがボールペンの値段、所持金の二つとなり、式をどのように立てようか迷う。そこで、考えてほしい式の作り方として、「分からないもの = 分からないもの」という作り方を試してほしい。
ボールペンの値段は文字で置いているので、所持金に関する式を立てることにする。

（ボールペン20本）
このとき、400円足りなかったので、所持金は
所持金 = 20 × X － 400　
　　　　　　　　　　となる。
（ボールペン15本）
このとき、50円足りなかったので、所持金は
所持金 = 15 × X －50
　　　　　　　　　　となる。
③式を解く
これまでに求めた式を使って、Xの値を求めていく。
15 × X －50 = 20 × X － 400
を解いて、X = 70円となる。

この中の②が一番のポイントです。まあ、そこが一番難しいんですけど、、、「わからないもの=わからないもの」の式の立て方は自分のレパートリーの中にあると、多くの問題を解くことができると思います。なので、何を文字で置いて、何を基にした式を立てるのかを意識しながら、問題に取り組んでみてください！検討を祈っています。1d:Td43,まずは数学の入試で求められる力が何かを考え入試で求められる力が何かを考えてみましょう。数学の入試で求められる力は主に3個あります。

1.『計算力』
これはいうまでもないでしょう。どれだけ正しいことを言っていても、またどれほど美しい解法を選択していたとしても途中で計算を間違えていれば、ほぼ0点になってしまいます。

2.『基本問題を覚えているかどうか』
ここでいう基本問題は青チャートのコンパス4個以下のもの、あるいは黄チャートに載っているすべての問題を指します。このレベルの問題で解けない問題が一つでもあれば難関大、ひいては東北大学の入試問題には太刀打ちできないと思います。

3.『基本問題を組み合わせられるか』
これがいわゆる「発想」と呼ばれる段階です。（今回は基礎に関する質問なので詳しくは述べません）

さて以上の3個のうち、基礎と言えるのは1と2の二つを合わせた力です。つまり、基礎が固まっている状態というのは、

「黄色チャートの問題すべてで瞬間的に解法が思いつき、計算ミスを一切すること計算ミスを一切することなくミスを一切することなく正解一切することなく正解に辿りすることなく正解に辿り着けること」

だということができます。

続いて効率的な基礎の固め方についてです。上に述べたことを踏まえれば、基礎を固める＝チャート式を完璧にすることです。なので僕がやっていた方法を紹介僕がやっていた方法を紹介します。
(Step1)とりあえず解いてみる。この時5分くらいは悩んでもいいと思います。答えが出なくてもOK
です。
(Step2)間違えたら→✖️、
　　　あってたけど次解ける自信がない→△
　　　あってて次も解けそう→◯
　　　のように印をつけて分類。
(Step3)上のStepを繰り返してとりあえず一周する（できるだけ短い時間で一周したい）
(Step4)×と△の問題のみ二周目をやる
(Step5~)上のことを繰り返す
これだけです。めっちゃ大変ですが、チャートは避けて通れません。上の方法を参考にしながらやって上の方法を参考にしながらやってみてください！ラストはこれが本当にできてるのかのチェックの仕方です。それはズバリ教科書傍用問題集を解くということです。解説がないからダメと言われていますが、基礎の確認にはもってこいなんですよね。レベルはチャートと同じで答えしかないわけチャートと同じで答えしかないわけですから、1と2のどちらかができていなかったら間違いということになります。つまり、自分に足りてないものが何かを正確に把握することができ正確に把握することができるわけ把握することができるわけです。一通りチャートをやったら傍用問題集をやってみましょう。復習にもなるし、計算力もつけられるしで一石二鳥です！以上です。参考にしていただければ幸いです。1e:Te99,こんにちは。
数学で応用問題を演習する時の話ですか？
それはズバリ演習量じゃないですかね
ただし、闇雲にやっても質は確保できません。
ではどうするか？今からまとめるのでこれを意識してください。

①使える定石は何か？
例えば、多変数関数の問題
まず変数同士の関係は従属なのか独立なのか？
また、独立なら初手の動きとしてどんな選択肢を取り得るか。こういうのを考えてください。
数学ができる人は1問から色んな学びを得ようとします。また、別解が思いつくような人はこういう思考をしていると思います。(別解がセンスだろってのもあると思いますがそういうのは除きます。ああいうのが思いつくのはほんとに演習量とセンスかなと。)

②15分手が動かなかったら答えを見る
この時間はさじ加減で決めてもらって構いませんが15分というのは今決めました。

そもそも難しい問題というのは多くの場合初手に何をすればいいか分からない、だったりこれで合ってるのか分からないというものばかりだと思います。なので大抵は最初に何をするかを決めるところが大事です。よってここで手が止まってしまうのなら、もうそれは演習不足によって思いつかないということなのではと思います。もちろんずっと考えていれば何か見えてくるものがあると思いますが、それはあくまで理想論です。

私も一時期わかるまで、わかるまでと頑張っていた時期がありましたがタイパが悪すぎてやめました。特に演習量が足りないうちは何していいか本当に分からないので、さっさと解答なりアプローチなりを見てほしいです。

③他教科とのバランスを考える
そんなの分かってるよと言われてしまうかもしれませんが一応の警告です。
あなたは新高校３年生だと思いますが、この時点でそこまで進んでいるなら数学はかなり順調と言えます。他教科もバランスよく進められているでしょうか？
本番でどの科目が難化しどの科目が易化するか分からないのでバランスが大事だと思います。
まだ共通テストのみの科目の対策をする必要があるかと言われたらまだいいんじゃないかとは思いますが二次科目は全て頑張って欲しいなと思います。

④定石をインプットする。
基礎固めは終わったって言ってるでしょ？て思ったかもしれませんが、応用問題も結局は基礎の積み重ねでしかないんです。(上述の通り、全て基礎ではなくある程度演習量等で決まってしまうものもありますが。)
方針で詰まることが多いということで、定石や基礎の部分が完璧では無いのかなというふうに思ってしまいました。青チャート等網羅系は本当に大事な事ばかり書いてあり、数学を本気でできるようにしたいならあれを完璧にするのが近道だと思っています。
私の他の質問者様への回答に高速周回なるものをご紹介していますので見ていただくとありがたいです。そんなに複雑なルールでは無いですし、なんなら自分好みにカスタマイズしてください。人それぞれ合う勉強法は違いますから。

こんな感じでしょうか？
質問等ありましたらお気軽にどうぞ。
応援していますっ！1f:Tb60,一言でいえば「経験」です。
必要条件の利用は青チャートなどのいわゆる網羅系参考書などでは得られない少し発展的な技術ですが、数学がある程度得意な人は過去にやったことがあったり、習ったことがあったりとそれを利用した経験があるのです。
なので質問者さんももう少し経験を積めば普通に利用できるようになると思います。

ここで必要条件の利用に至るまでの思考回路の簡単な例を紹介しておきます。
問題
「k を正の整数とする. 5n^2 − 2kn + 1 < 0 ー①を満たす整数 n が，ちょうど 1 個であるような k をすべて求めよ.」
これは2008年の一橋の問題です。下にプロセスを書いてますが良問であり考えがいがあるので一回自力で考えてみてください。


まず大前提として数学における問題と解は全て必要十分性を保っている必要があります、従ってこの問題を解く際必要十分を保ちながら解く(同値変形)のと必要、十分を分けて解く2通りに分かれます。この問題では実数でなく整数の2次方程式であり同値変形で解くのはややこしい(できることはできます)と判断しまず必要条件から絞ろうと考えます。
この問題を考える際式①が成り立つとき5x^2-2kx+1=0(xは実数)が二つの異なる実数解を持つー②「必要」がある(つまり②は①の必要条件である)ことを利用します、そうすることによってkの条件が分かります。そのもとで①を満たす整数nがちょうど1個である条件は5x^2-2kx+1=0の二つの解(α、βとおく)の差が2未満ー③であればいいということがわかります。②と③から得られるkの範囲が5=<k^2=<30かつkは正の整数よりk=3,4,5であることがあることが必要。ということがわかります。これらを実際に①に代入し十分性を確認することで必要十分性が保たれるわけです。(解はk=4,5です)今は難しいかもしれませんが必要条件を使う問題にたくさん触れることでスッと理解できるようになると思います。また僕が思いつく限りでもあと2つ別解があるので3年生になってからでも良いので試してください。

また全てnについて(n,a,xについての式)がなり立つためのaの条件を求めよなどの問題でも全てのnってことはn=1,2でも成り立つ「必要」があるな、と思い試したりすることもよくあるので覚えといてください。

質問者さんはまだ高1ということで完全に理解することはむずかもしれませんが受験期になればきっと理解できると思うのでそれまで勉強に励んでください。2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=9cOnEx4keDeHtVM7FYx8","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"夏休みの課題プリント解説"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",1,") コメント(",3,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"8/23 17:11"}]]}],["$","div",null,{"className":"coach-mark 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mb-4","children":"すべての回答者は、学生証などを使用してUniLinkによって審査された東大・京大・慶應・早稲田・一橋・東工大・旧帝大のいずれかに所属する現役難関大生です。加えて、実際の回答をUniLinkが確認して一定の水準をクリアした合格者だけが登録できる仕組みとなっています。"}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap mb-4","children":[["$","div","advice-part-0",{"children":[null,"立方体を一個作る時に棒は12本必要です。つまり純粋に立方体を2個作るには24本要ります。しかし立方体の面と面をくっつけた時4本分が相殺されるため、一個繋げる時12本必要で、二個繋げる時は20本、三個繋げる時は28本必要となります。こう考えると、式は8X＋4となります。"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":null,"adviserName":"Polo","adviserDepartment":"大阪大学外国語学部","adviceId":"9cOnEx4keDeHtVM7FYx8","numberOfFan":0,"clipsAvg":1,"adviceRateAvg":5,"profile":""}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink パンフレットバナー画像","className":"mt-4 rounded"}]}]}],["$","div",null,{"className":"pt-4","children":["$","$L17",null,{"id":"adsbygoogle-init-under-advice"}]}]]}],["$","div",null,{"className":"flex justify-between","children":[["$","h1",null,{"className":"text-xl font-semibold","children":["コメント(",3,")"]}],["$","$L18",null,{"adviceId":"9cOnEx4keDeHtVM7FYx8"}]]}],["$","div",null,{"className":"mb-8","children":["$","div",null,{"className":"divide-y","children":[["$","div",null,{"className":"flex py-4","children":[["$","div",null,{"className":"mr-2","children":["$","$L19",null,{"avatarUrl":null,"contributorName":"ダンゴムシ","adviceId":"9cOnEx4keDeHtVM7FYx8"}]}],["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"flex justify-between","children":[["$","div",null,{"className":"mb-2","children":["$","$L1a",null,{"contributorName":"ダンゴムシ","adviceId":"9cOnEx4keDeHtVM7FYx8"}]}],["$","div",null,{"className":"text-xs text-caption","children":"8/23 17:53"}]]}],["$","div",null,{"className":"text-xs whitespace-pre-wrap","children":"分かりやすい説明ありがとうございます！"}]]}]]}],["$","div",null,{"className":"flex py-4","children":[["$","div",null,{"className":"mr-2","children":["$","$L19",null,{"avatarUrl":null,"contributorName":"ダンゴムシ","adviceId":"9cOnEx4keDeHtVM7FYx8"}]}],["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"flex justify-between","children":[["$","div",null,{"className":"mb-2","children":["$","$L1a",null,{"contributorName":"ダンゴムシ","adviceId":"9cOnEx4keDeHtVM7FYx8"}]}],["$","div",null,{"className":"text-xs text-caption","children":"8/23 19:50"}]]}],["$","div",null,{"className":"text-xs whitespace-pre-wrap","children":"8X+5の8と4は、どこから来ましたか？"}]]}]]}],["$","div",null,{"className":"flex py-4","children":[["$","div",null,{"className":"mr-2","children":["$","$L19",null,{"avatarUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_mBLd8WqR2pOVKp8vWtIyDjP2qA03.jpg?alt=media&token=58f8baf6-452b-4be3-b749-a23ab3e01e88","contributorName":"たけなわ","adviceId":"9cOnEx4keDeHtVM7FYx8"}]}],["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"flex justify-between","children":[["$","div",null,{"className":"mb-2","children":["$","$L1a",null,{"contributorName":"たけなわ","adviceId":"9cOnEx4keDeHtVM7FYx8"}]}],["$","div",null,{"className":"text-xs text-caption","children":"8/23 23:31"}]]}],["$","div",null,{"className":"text-xs whitespace-pre-wrap","children":"回答者ではございませんが、横槍失礼致します。棒12本で立方体1個なので、立方体X個で考えれば12Xとなりますが、接着面においては4本の棒が重複しているのでこれを引く必要があり、立方体をX個繋げたときの接着面の数は全部で（X－1）個なので、4は（X－1）回引くことになります。したがって、\n12X－4（X－1）＝8X＋4\nとなります。"}]]}]]}]]}]}],["$","h1",null,{"className":"text-xl font-semibold","children":"よく一緒に読まれている人気の回答"}],["$","div",null,{"className":"mb-8","children":["$","div",null,{"className":"divide-y","children":[["$","div",null,{"children":["$","$L7",null,{"href":"/advice/kYFt1poHhTM0MEnd6H8V","children":["$","div",null,{"className":"flex items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"質問 何を数え落とした"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"かきふらいさん初めまして。\n文系ですが、回答させていただきます。\n\n本回答の構成は以下です。\n①数え落とした5通りはなにか\n②再考察が合っているか\n③数え上げる際のポイント\n\n以上3点です。\n\n①数え落とした5通りはなにか\n結論としては、端が白玉になる場合が想定されていないかなと思います。\n\nex.●◯●●◯●●◯●●◯●◯●◯\n\n上記のような場合です。\n\n②再考察が合っているか\n結論、誤りです。\n本問の場合、同色の玉に区別は存在しません。\n\n区別する場合は玉に番号が振ってあったりする場合でしょう。本問のように、玉を「色」でしか判別していない場合、区別する必要がないのです。\n\n③数え上げる際のポイント\n質問が若干抽象的なので、回答者様の解法に近いもので解く場合、私なら以下のように解きます。\n\nまずは、解答のための必要条件を具体的に導きます。\n必要条件は、\n「白玉同士は隣接しない」\n「赤玉は白玉に隣接する」\n以上の2つです。\n\nしたがって、初期配置は以下のようになります。\n\n◯●◯●◯●◯●◯●◯\n\n上記のように、赤5個、白6個を配置できます。残る赤4個をどこに配置するかですが、以下のようになります。\n\n1◯●2◯●3◯●4◯●5◯●6◯7\n\nつまり7箇所、赤を置く場所があります。なお、１つの場所につき置けるのは1つまでです。一見、「◯●」の間に置く場合も考えた方が良さそうな気がしますが、これは考慮する必要がありません。なぜなら、先ほど述べた通り、同色の玉はそれを区別する必要がないからです。\n例えば、\n「1◯●2」の2に赤を置けば、「1◯●●」となります。\n次に、\n「1◯●2」の◯●の間に●を置く場合を想定しても「1◯●●」となります。\nこのように、結果が同じになる事が分かるかと思います。\n\nしたがって、7箇所のスペースに4つの赤を置くケースを想定するので、\n\n7C4＝35\n\nよって、35通りとなります。"}],["$","div",null,{"className":"flex 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"この問題教えてください！"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"質問に対して結論から言うと、2^n+2で割るのも大丈夫です！\n\n　解説が2^n+1で割ると言うように書いているのは、左辺をAn+1/2^n+1のように綺麗にまとめられるからだと思われます。そのあとは、Anと2のべき乗のnが対応するように整理して解き進めていくとex)An+1/2^n+1 、どちらの方法でも同じ形になることがわかると思います。\n\n　私なりに解答の思考プロセスがどのようなものかご説明すると、「An/2^nの形を作りたいから、左辺で一旦綺麗にまとめてみよう！」と言った感じです。\n\n　質問者様のように、2のn乗を消すことから始めるのも、定石に則っていて素晴らしいと思います。\n\n　最後に総括すると、どのような式変形をしていきたいのかをまず考えて、逆算的に解き進めていくことが大切にしていく必要があるということですね。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 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