Tính.
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
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Sửa đề: \(\left(4x^4+14x^3-21x-9\right):\left(2x^2-3\right)\)
\(=\left(4x^4+14x^3+6x^2-6x^2-21x-9\right):\left(2x^2-3\right)\)
\(=\left[\left(4x^4-6x^2\right)+\left(14x^3-21x\right)+\left(6x^2-9\right)\right]:\left(2x^2-3\right)\)
\(=\left[2x^2.\left(2x^2-3\right)+7x.\left(2x^2-3\right)+3.\left(2x^2-3\right)\right]:\left(2x^2-3\right)\)
\(=\left(2x^2+7x+3\right).\left(2x^2-3\right):\left(2x^2-3\right)\)
\(=2x^2+7x+3\)
___________________
\(\left(6x^3-2x^2-9x+3\right):\left(3x-1\right)\)
\(=\left[\left(6x^3-2x^2\right)-\left(9x-3\right)\right]:\left(3x-1\right)\)
\(=\left[2x^2.\left(3x-1\right)-3.\left(3x-1\right)\right]:\left(3x-1\right)\)
\(=\left(2x^2-3\right).\left(3x-1\right):\left(3x-1\right)\)
\(=2x^2-3\)
`#NqHahh`
\(\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{4}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{4}{13}\right)\)
\(=\dfrac{6}{7}\cdot\dfrac{13}{13}=\dfrac{6}{7}\)
\(\dfrac{6}{7}\).\(\dfrac{8}{13}\) + \(\dfrac{6}{13}\).\(\dfrac{9}{7}\) - \(\dfrac{4}{13}\).\(\dfrac{6}{7}\)
= \(\dfrac{6}{7}\).\(\dfrac{8}{13}\) + \(\dfrac{6}{7}\).\(\dfrac{9}{13}\) - \(\dfrac{4}{13}\).\(\dfrac{6}{7}\)
= \(\dfrac{6}{7}\).(\(\dfrac{8}{13}\) + \(\dfrac{9}{13}\) - \(\dfrac{4}{13}\))
= \(\dfrac{6}{7}\).1
= \(\dfrac{6}{7}\)
Olm chào em, Cảm ơn em đã phản hồi tới olm. Tất cả các câu hỏi linh tinh đều đã bị cô xử lí theo hình thức phù hợp với mức vi phạm, nhẹ thì trừ gp, nặng thì khóa tài khoản em nhé.
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
=>\(3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
=>\(3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)
=>\(2B=1-\dfrac{1}{3^{100}}=\dfrac{3^{100}-1}{3^{100}}\)
=>\(B=\dfrac{3^{100}-1}{3^{100}\cdot2}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}\\ \Rightarrow2B=\dfrac{3^{100}-1}{3^{100}}\\ \Rightarrow B=\dfrac{3^{100}-1}{2\cdot3^{100}}\)Vậy \(B=\dfrac{3^{100}-1}{2\cdot3^{100}}\)