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Therefore the area $(a+b)^{2}$ of $S$ equals the sum $a^{2} + b^{2}$ of the areas of $S'$ and $S''$ plus the sum $ab + ab = 2ab$ of the areas of $R$ and $R'$, i.e. $$(a+b)^{2} = a^{2} + b^{2} + 2ab.$$

2024年10月22日 · I'm going through Cunningham's A Logical Introduction to Proof. An exercise is asking to prove that $a^2+b^2 > 2ab$, by letting $a$ and $b$ be distinct numbers and using the fact $x^2 >0$ I

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2019年3月3日 · det(A + B)2 = 8 det(A2 +B2). My work: Let A = X + Y, B = X − Y with X, Y ∈M3(C). The condition rewrites as −2Y2 = [X, Y]. Since Y commutes with Y2 it will also commute with [X, Y], so according to Jacobson's lemma [X, Y] is nilpotent.

Factor using the perfect square trinomial rule a2 + 2ab+b2 = (a+b)2 a 2 + 2 a b + b 2 = (a + b) 2, where a = a a = a and b = b b = b. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a …

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Raise 2 2 to the power of 5 5. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.