(3x+1)^2-2(3x+1)(3x+5)+(3x+5)^2 (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1) 24/08/2021 Bởi Piper (3x+1)^2-2(3x+1)(3x+5)+(3x+5)^2 (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
Đáp án: \(a,\ 16\\ b,\ \dfrac{3^{64}-1}{2}\) Giải thích các bước giải: \(a,\ (3x+1)^{2}-2(3x+1)(3x+5)+(3x+5)^{2}\\ =[(3x+1)-(3x+5)]^{2}\\ =(3x+1-3x-5)^{2}\\ =(-4)^{2}\\ =16\\ b,\ (3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ \text{Đặt A = $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)$}\\ \Rightarrow (3-1)A=(3-1)(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{2}-1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{4}-1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{8}-1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{16}-1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{32}-1)(3^{32}+1)\\ ⇔2A=3^{64}-1\\ ⇔A=\dfrac{3^{64}-1}{2}\\ \text{Vậy biểu thức bằng $\dfrac{3^{64}-1}{2}$}\) chúc em học tốt! Bình luận
Đáp án:
\(a,\ 16\\ b,\ \dfrac{3^{64}-1}{2}\)
Giải thích các bước giải:
\(a,\ (3x+1)^{2}-2(3x+1)(3x+5)+(3x+5)^{2}\\ =[(3x+1)-(3x+5)]^{2}\\ =(3x+1-3x-5)^{2}\\ =(-4)^{2}\\ =16\\ b,\ (3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ \text{Đặt A = $(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)$}\\ \Rightarrow (3-1)A=(3-1)(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{2}-1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{4}-1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{8}-1)(3^{8}+1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{16}-1)(3^{16}+1)(3^{32}+1)\\ ⇔2A=(3^{32}-1)(3^{32}+1)\\ ⇔2A=3^{64}-1\\ ⇔A=\dfrac{3^{64}-1}{2}\\ \text{Vậy biểu thức bằng $\dfrac{3^{64}-1}{2}$}\)
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