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1c:Te31,こんにちは！
たしかに三元一次は煩雑になってミスりがちですね笑
自分もベクトルの大きさの計算なんかはかなり苦手でした。

以下、計算ミスを防ぐために（特に共通テストで）気をつけるポイントをお伝えします！



①ベクトルの成分は縦に書く
もしかしたら既にやっているかもしれませんが、ベクトルの成分は縦に並べて書きましょう。現行の教科書などは成分が横『（２，4，3）のような形』で書かれていることが多いですが、これだとミスりやすいです。
　　2
｛　4　｝
　　1
のように縦で成分表示すると文字が入って式が複雑になっても見やすいので、成分同士の方程式や内積の計算をするときのミスがかなり減ります。

(OP→)＝x(a→) ＋y(b→) + z(c→)
のような場合も、
　　　　　　ａ
(OP→)＝｛　ｂ　｝
　　　　　　ｃ
のように表しちゃうと計算でミスりづらいです！


②大きな余白や白紙のページを利用する
共通テスト本番ではめちゃくちゃ煩雑なベクトルの計算が出ることは正直あまりないです。しかし、東進などの予備校が手掛けている模試や問題集の中には、計算ゲーのような悪問も含まれているのが現状です。ですので正直に言えば、そういった模試などの悪問でケアレスミスをしてしまっても一喜一憂することは無いと思います。
しかし、工夫をするとすればやはり余白の使い方でしょう。「あ、この計算重いわ」と感じたら、無理して小さい余白や暗算に頼らず、どっしりと構えて大きな余白を探しましょう。その分タイムロスに感じるかもしれませんが、いくらわさんのように京大を目指すレベルであれば、タイムロスよりも安易な判断による失点の方が痛いことは明確だと思います。心に余裕を持って頑張ってください！


③後回しにする
共通テストの数学は、ひらめきゲー／誘導ゲーな要素があります。自分のやり方でやったら死ぬほど難しい式がでてきたけど，誘導にうまく乗っかって解き直したらめちゃくちゃ簡単だった、なんてケースがかなり多いです。また、わからないからとりあえず飛ばして最後に戻ってきたら、頭がクリアになって簡単に解けたというケースも多いです。
問題が変に難しいなと感じた時は、割り切ってスキップして、最後に戻ってくるようにしましょう。仮に計算ミスをしていたとしても、後で見直すと間違いに気づきやすいです。共通テストはとにかく時間と勝負なので、沼りはじめたら終わります。とりあえずスキップしてみることは案外大切な心構えですよ！


①〜③までご紹介しましたが、特に大事なのは③です。
これは共通テストの数学では本当に大切な考え方です！一緒に受験勉強していた東大生の友人たちでさえ、計算が煩雑になったり沼ったりすることがありましたし、そういう時はとりあえず飛ばして最後に戻ってくるのがいいと話していました。

ぜひ参考にしてください！
また、これから過去問などで形式に慣れていけば、だんだん計算ミスは減ってくると思いますよ〜！頑張ってください！1d:T1208,センター英語は勉強の仕方が多様にあります。
私としては大問ごとに練習するのが一番だと思います。
苦手な大問だけを過去問から引っ張ってきて、連続で１日のうちに演習しまくれば自然と力はつきます。あ、これは一般入試の過去問でも同様です。

センター英語でまずやらなくてはならないのは、大問２を１つも落とさないことです。これから全ては始まります。文法問題は基本的な四択で、その後に並び替えがありますが、インプット知識的に満点取れるレベルなので落としてはいけません。

で、その後に大問５、6を１つも落とさないような気持ちで練習をします。5は小説か日記で、そこそこ難しいです。昨年のオクトパスの設問は結構話題になりましたが、本文に直接的にoctopusとかかれてなくても、eight legsといった表現から宇宙人がタコであると理解しなくてはなりませんでした。
設問にsuggestとあったら、本文にはないがヒントはあるかもしれない設問だと認識しておくといいでしょう。これは大問４のAの最後の設問で、続きのパラグラフの内容を答える設問がありますが、最終パラグラフの内容から推察する必要がある問題です。
大問6に関してはパラグラフ毎の内容を問う問題なので、落とせません。

さて、2.5.6と話を進めたら今度は3番です
3番Aは不要文削除で苦手な人が多い大問ですが、ここは自分は大得意でした。

そのカラクリを教えます。

まず、よく考えてみてください。
あんなに小さな段落に内容を１つ詰め込むとなると、一文の情報はものすごく密になるはずです。

ちなみに国公立の英文が短いくせに読みづらいのに対し、超長文は長い割に読みやすいものです。

内容の量は英文１つで変わりはしません。
変わるのは、長ければ長いほど具体例とかが入れられるのでわかりやすくなるのです。
短ければ短いほど名詞構文「the analysis of this project is ,,,」みたいなものが増え、一文にめちゃくちゃ詰め込むからわかりづらいのです。
しかも文と文の関係性がより密になるので、最初のほうがわからなくなると、ドミノ倒しで全部わからなくなるのです。

これを逆手にとると、文と文の関係性が密になるということで、①から④の文は
⑴前後の設問になっていない文との関係
⑵設問になってる①から④の文同士の関係
この２つの関係がないものが不要な文なのです。

この大問はパターンがわりと決まっているので、演習量が物を言います。ぜひ、連続で演習してみてください。

もう一つの大問3のBは会話文ですが、会話文は質問に対して解答している部分を要約しろという形態になっています。最初の２つはそのまま要約すればいいだけですが、最後のは伏線回収的な設問となっているので、最初のアンサーの人から最後のアンサーの人までの共通テーマを見いだしながら読んでいくと効率的です。

そして大問４です。
大問４A自体は読めば解ける設問なのでとくにはありませんが、最後の設問は前述した通り、気をつけて。

大問4Bは9割のるか乗らないかの設問です。
最初に設問に目を通して、問われてることを確認してからチラシを隅々まで点検しましょう。先にチラシみると見るべきポイントがわからないのでしっかりやってください。

ここまでで1以外は説明しました。
１は満点ゲームでキモになるだけで
正直９割とるには不要です。あとの部分でどれだけ稼げるかです。

さて、モチベーション的な話にします。

最初の話とは矛盾してしまうかもしれませんか
最終形態では、1、2は落としてもいいので3.4.5.6でミスを防いでください。
1,2に比べて3,4,5,6は一問の配点が高く、落とすとキツイです。１、２は落としても2点なので、あまり痛くありません。
最後は3.4.5.6が完璧になるよう学習を進めていくといいでしょう。

長くなりましたが、センター攻略法でした。

1e:T149c,こんにちは。共テ数学は問題構成が特徴的な問題が多いので、記述式の問題とは少し違った解き方をしないといけないと思います。
全統の数学で偏差値65が取れているということは基礎が概ね固まっていると思うので、共テ数学をスムーズに解くコツさえ習得できれば大丈夫だと思います。以下に共テ数学に臨む上で念頭に置いた方が良いと思った事項をまとめました。

◎共テ数学は「数学」ではなく「情報処理」
まず、共テ数学は「情報処理」を高速で繰り返すものだという認識をもっていただけると良いと思います。残念ながら、一つ一つの問題をじっくり吟味する時間はありません。解く時間が足りない、というのは、最初から一つ一つの問題に丁寧に向き合ってしまっているからではないかと推測しています。もしそうであれば、もう少しラフに問題に立ち向かっても良いかと思います。例えば、ある関数f(x)が最初に定義されていて、唐突に「f(2)を求めてください」と言われていたら「なぜx=2をここで代入するんだろう」と考えるのではなく、直ちに何も考えずにx=2を代入して計算してみてください。「なぜそのような操作が必要なのか」を考える前にとりあえず計算してみる、ということが非常に重要だと思います。一つの問題に対して様々な解法が存在することは多々あり、共テ数学はマーク式なのでその解き方が限定されてしまっています。自分の思いついた解きやすい方法とは違い、一見遠回りに見える方法で解くように指示されていることも少なくはないでしょう。そのような問題構成になっている以上、「解法を理解してから解き始める」というのはどうしても時間がかかってしまいます。とりあえず言われた通りに値を求めて手を動かしていくと、作問者が提示した解法の意味がだんだん理解できてくることが増え、スムーズに問題を解いていくことができると思います。

◎必要な情報を素早く見つける
上述したように、共テ数学の問題では「なぜ今これを求めなければならないのか」と思うような問題があります。突然何の脈絡もなく値を求めさせるような問題がポンと出てきた時は特に、後の問題でその値を利用するものだと思って解き進めてみてください。直後の問題だったり、次の小問だったりと、多少の差はありますが、その答えを利用する問題が後ろにきっと出てくるはずです。問題文に含まれている不可解な情報についても、「問題文中に意味のない情報は含まれていない」と思って、ただその時はあまり深く考えずに、頭の片隅にそれを入れておくことが重要だと思います。解き進めていくうちに詰まってしまった時は、そのことを思い出して前に戻り、それらがうまく利用できないかを考えてみると打開できることがあると思います。

◎誘導がなくなったら「前の小問の解法･求めた値を利用」すべし
大問の中で(1)、(2)、(3)とあった場合、(3)に誘導がほとんどついていないことが多いと思います。そういう時は、やはり前の小問に戻って解法を確認し、同じ要領で答えを導き出す、という方法が一番早いことが多いです。また、場合の数･確率の問題などでは、解法だけでなく求めた値も利用すると計算が楽になることもあります。例えば過去に見たことのある問題では、はじめ二種類のカードを無作為に引いて並べ、その並び方を調べる、という小問からスタートし、その後カードを三種類に増やし、引く枚数を4枚、5枚と増やしていくものがありました。そのような問題では、カードが1枚ずつ増えるときは「n枚がn+1枚になったんだから、n枚について考えた前問の答えを使えば追加した1枚についてだけ考えれば良い」という発想で楽に解くことができるようになります。ただし、前問の答えを利用する場合、計算ミスによる雪崩にはくれぐれも気をつけていただきたいと思います。

◎まとめ
もう一度上述した内容をまとめると、

･とりあえず言われた通りに計算してみる
･不可解な情報、「なぜここでこの値を求めるのか」に要注意
･誘導が消えたら前問の解法をチェック、必要ならば前問で求めた値を利用

の三つを念頭に置きながら解くことが重要だと思います。繰り返しますが、共テ数学は「情報処理」です。記述式の問題とは種類が違う、ということを踏まえ、問題に取り組んでみてはいかがでしょうか。少しでも参考になれば幸いです。1f:Tc0d,お答えしますね！
僕はミスが絶えないタイプの人間だったのですが、それを克服して、本番は一切計算ミスなどをすることなく合格することができました。ミスを減らす事は合格に直結するので僕のやったことを書いてみます。

さて、そもそもなぜミスするのでしょうか。もちろん注意力不足というのはあるでしょう。（この減らし方は後で書きます）ですが真っ先に疑って欲しいのは学力不足です。人間は自分のレベル以上のことをやろうとすると、いろいろなことを考えるためにミスが増えます。逆に自分にとって簡単なことはまずミスをしません。例えば九九や一桁の数字の足し算などです。これらをミスすることがあるでしょうか。おそらくないと思います。それは僕らのレベルがそのレベルの計算をはるかに凌駕しているからです。一方、積分などを習いたての頃は、それなりに苦労をしたと思います。それは自分のレベルのギリギリでものを考えているからです。後から見れば「符号を間違えた」という程度のことであっても、それが演習不足、経験不足に起因する事は多いです。自分の学力を疑うことは辛い事ですが、ここと向き合うことで今以上に実力を伸ばすことができると思います。

続いて、純粋なケアレスミスの減らし方です。学力が十分にあっても、間違えるときは間違えます。それを減らすにはどうすればいいでしょうか。その方法の一つとして
「自分のミスを分析したノートを作り、問題を解くたびにそのノートを見返す」
というのがあります。実は自分がやるミスというのは限られています。自分がどういうミスをした自分がどういうミスをしたのかをノートにまとめ、いつでも見返せるようにしておくと、問題を解くときにミスしないように注意すべきポイントがわかり、格段にミスが減ります。ぜひやってみてください。

さて最後に、「計算結果を覚えてしまう」という方法があります。なぜミスをするのかというと、そもそも計算するからです。計算しなければミスのしようがないわけです。なぜ九九を間違えないかといえば、それは九九を覚えているからです。なので、30²以下の平方数（16²=256など）などよく使う計算は結果を含めて覚えておくというのも一つの手です。

以上です。少し厳しいことも言いましたが、「数学でミスをする＝30点がなくなる」です。どれだけ考え方があっていても、計算結果が間違っていればほとんどの場合点はありません。それくらいシビアなものだと思って、ミスをなくせるように頑張ってください！2:["$","main",null,{"className":"px-4 pt-4 pb-4","children":["$","div",null,{"className":"max-w-3xl mx-auto w-full","children":[["$","div",null,{"className":"mb-8","children":["$","$L7",null,{"href":"https://unilink-app.onelink.me/isbO/h6xeh63x?advice=ocdjlWUBVTfCb-RZpaUZ","target":"_blank","children":["$","$L8",null,{"src":"/images/web_to_app_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink WebToAppバナー画像","className":"mb-4 rounded"}]}]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"重複組合せの公式が意味不明"}],["$","div",null,{"className":"flex justify-between mb-4","children":[["$","div",null,{"className":"text-left text-xs text-caption","children":["クリップ(",2,") コメント(",1,")"]}],["$","div",null,{"className":"text-right text-xs text-caption","children":"9/1 22:47"}]]}],["$","div",null,{"className":"coach-mark mb-4","children":"UniLink利用者の80%以上は、難関大学を志望する受験生です。これまでのデータから、偏差値の高いユーザーほど毎日UniLinkアプリを起動することが分かっています。"}],["$","div",null,{"className":"mb-4","children":["$","$L13",null,{"clientImageUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_gqJZ1R84SHRtqydT.jpg?alt=media&token=82c144b0-8aa1-44c4-b465-cfbcef3f826d","clientUserName":"ともくん","infoString":"高3 埼玉県 一橋大学経済学部(70)志望","adviceId":"ocdjlWUBVTfCb-RZpaUZ"}]}],["$","div",null,{"className":"mb-8","children":[["$","div",null,{"className":"leading-loose whitespace-pre-wrap","children":[["$","div","consultation-part-0",{"children":[null,"数Ａの場合の数の重複を許して作る組合せからです。\n例えば1〜6までの6つの数字から重複を許して3つの数字を選ぶ、です。\nなぜ異なるｎ個のものから重複を許してｒ個取る組合せの総数は「n  r−1Ｃr」となるのですか？\n分かる方宜しくお願いします。"]}]]}],["$","div",null,{"className":"pt-4","children":["$","$L14",null,{}]}],null]}],["$","h1",null,{"className":"text-xl font-semibold mb-2","children":"回答"}],["$","div",null,{"className":"mb-4","children":["$","$L15",null,{"adviserImageUrl":null,"adviserName":"kmhr","adviserDepartment":"東京大学理科二類","adviceId":"ocdjlWUBVTfCb-RZpaUZ"}]}],["$","div",null,{"className":"coach-mark 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":[["$","div","advice-part-0",{"children":[null,"n>rの方が分かりやすいので、1〜3の数字から重複を許して6つ選ぶとします。\n◯◯|◯|◯◯◯\nのように6つの丸◯を2本の棒|で仕切ることを考えてください\nそして一番左の区切りにある◯が1、真ん中が2,右が3とすると、\n112333\nとなりますね\n例えば2が一度も使われない場合、\n◯◯||◯◯◯◯\nとなります。\nつまり、1〜3の数字を重複を許して6つ選ぶ場合の数は、◯4つと|2本の並べ方の場合の数になります。\nさて、1から6の数字から重複を許して3つ選ぶ場合、\n例えば\n◯|||◯||◯\nのように、３つの◯を5本の|で仕切る、つまり3つの◯と5本の|の並べ方の場合の数になりますね"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":null,"adviserName":"kmhr","adviserDepartment":"東京大学理科二類","adviceId":"ocdjlWUBVTfCb-RZpaUZ","numberOfFan":4,"clipsAvg":2.7333333333333334,"adviceRateAvg":4.625,"profile":"東京大学大学院理学系研究科所属です。中部地方出身、中高一貫卒です。"}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink 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7\nとすることですべての自然数を表すことができます。問題で聞いているのは平方数ということなので、それぞれを2乗すると、\n\n0→64n^2=8×8n^2\n1→64n^2 16n 1=8(8n^2 2n) 1\n2→64n^2 32n 4=8(8n^2 4n) 4\n3→64n^2 48n 9=8(8n^2 6n 1) 1\n4→64n^2 64n 16=8(8n^2 8n 2)\n5→64n^2 80n 25=8(8n^2 10n 3) 1\n6→64n^2 96n 36=8(8n^2 12n 4) 4\n7→64n^2 112n 49=8(8n^2 14n 6) 1\n\nとなります。\nすべて(8n ○)^2という式になる以上、n^2とnの係数は8の倍数になるので、自然数部分である余りの2乗部分を8で割った時の余りが平方数の余りになります。\n\n長くなってすみません。わからなかったらまた質問してください。"}],["$","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 3.31-2 6-2m0-12C9.79 4 8 5.79 8 8s1.79 4 4 4 4-1.79 4-4-1.79-4-4-4zm0 10c-2.67 0-8 1.34-8 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items-center py-4","children":[["$","div",null,{"className":"flex-1","children":[["$","div",null,{"className":"mb-1","children":"京大数学はじめの全問題触る時間"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"こんばんは、さかさかです。\n\n\n時間配分ってなかなか難しいですよね。私が解いていた時は、10分から20分くらいを目安に解く問題をきめていました。\n\n合格ラインが3完と少しくらいで完璧に全ての問題を解ける必要はないので、解ける問題をしっかりと選別するのが1番大事になってきます。なので解く時間は少し減ってしまいますが、選別に時間を多めにかけた方がいいかなと思います！\n\n体感ですが京大の理系数学は6問のうち、解けそうな問題、頑張ればいけるかな、、？な問題、かなり難しい問題、がそれぞれ2問ずつあるイメージです。\n\nただ最初に軽く触れただけで分別するのは難しいので、私はまず15分ほどかけて、少しでも方針が立つ問題と全く立たない問題の二つに大別していました。その後の具体的な時間配分としては、\n\n仮に、少しでも方針が立った問題が3問、全く方針が立たない問題が3問あったとすると、先に前者の3問を時間をかけて完答を目指して解きます。一問に25分ずつかけて解いたとして、2問が完答、1問が0.5完答になったとします。\n\nこの時点で15＋25×3＝90分が経過していて残りが60分あるかないかで、そこから残りの3問を解いていきます。順番はランダムにとりあえず手をつけてみて、10分立っても方針がつかなかった場合は飛ばして次の問題に取り掛かるという感じで残りの問題は進めていました。\n3.5完を目指すなら残り1完、あわよくば1.5完ができるのといいのかなと思います！\n\n\nあくまでも私の時間配分の例ですが、参考にしていただけると幸いです☺️\n他に気になることがあれば、気軽にコメント欄で聞いてください！\n\n\n\n\n\n\n"}],["$","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":"文法問題や並べ替えで間違える"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"慶應の経済学部の者です。\n\n並べ替え問題ができない理由は主に2つ挙げられます。\n\n❶文法事項が有機的に結びついていない。\n問われている文法事項が１つだと答えられるのに、問われることが2つや3つになると出来ないなんてことありませんか？\nこれは❷にも繋がる話なんですが、並べ替えで組み合わせる文法事項の組み合わせなどをきっちり個々に理解しておく必要があります。\n苦手な事項等があれば早急に理解して、問題も解けるようにしてください。\n\n❷並べ替えの経験不足\nだいたい並べ替え問題で問われることなんて決まってます。\n問題数をこなせば『あ〜あのパターンね』みたいに問題か解けるようになります。\n場合によっては単語見ただけですぐ並べ替えのパターンが見えることだってありまふ笑\n漫然と問題を解かずに、しっかり何が問われているかを意識しながら問題を解いてみてください。\n\n僕からのアドバイスはこれくらいです。\n是非意識してみてください！\n応援しています！頑張ってください！"}],["$","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 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