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Home » \frac1n(n+2) = \fracAn + \fracBn+2. - 500apps

\frac1n(n+2) = \fracAn + \fracBn+2. - 500apps

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 بعد ذلك، نحتاج إلى اختيار 3 أوراق غير آسك من البقايا الـ48 ورقة (بما أن 52 - 4 = 48): تُعطى هذه بـ \(\binom{48}{3}\):

📌 \binom{48}{3} = \frac{48 \times 47 \times 46}{3 \times 2 \times 1} = 17296

📌 إجمالي عدد اليد الفريدة المكونة من 5 أوراق تحتوي على ورقتين آس فقط هو حاصل ضرب هذه القيم:

📌 \times 17296 = 103776

📌 إذن، عدد مثل هذه اليد هو \(\boxed{103776}\).問題: 不等式 \(3x^2 - 12x + 9 \leq 0\) を満たす \(x\) の最大値を求めよ。

📌 解: まず、関連する二次方程式 \(3x^2 - 12x + 9 = 0\) を解いて、臨界点を求める。簡略化すると \(x^2 - 4x + 3 = 0\) となる。これは \((x-1)(x-3) = 0\) と因数分解でき、\(x = 1\) と \(x = 3\) が根である。これらの根は数直線を区間に分け Preparing intervals: \((- \infty, 1)\), \((1, 3)\), \((3, \infty)\) を定める。各区間において式の符号をテストする:

📌 \(x \in (-\infty, 1)\) の場合、\(x = 0\) をとぶ: \(3(0)^2 - 12(0) + 9 = 9 > 0\)。

📌 \(x \in (1, 3)\) の場合、\(x = 2\) をとぶ: \(3(2)^2 - 12(2) + 9 = -3 \leq 0\)。

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