TL nhanh giúp.mình ạ
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\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
\(A:B=\left(3x^4+3x^2+x^3+x-3x^2-3+5x+8\right):\left(x^2+1\right)\\ =\left[3x^2\left(x^2+1\right)+x\left(x^2+1\right)-3\left(x^2+1\right)+5x+8\right]:\left(x^2+1\right)\\ =3x^2+x-3\left(\text{dư }5x+8\right)\\ \text{Vậy }A=\left(x^2+1\right)\left(3x^2+x-3\right)+5x+8\)
a, \(A=-\dfrac{1}{20}-\left(\dfrac{1}{20\cdot19}+\dfrac{1}{19\cdot18}+...+\dfrac{1}{2\cdot1}\right)\\ \Rightarrow A=-\dfrac{1}{20}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\\ \Rightarrow A=-\dfrac{1}{20}-1+\dfrac{1}{20}=-1\)
b, \(B=\dfrac{1}{99}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ \Rightarrow B=\dfrac{1}{99}-\dfrac{1}{2}+\dfrac{1}{2\cdot99}=-\dfrac{16}{33}\)
3 + 4 - 1 + 2 = 8
Mình cx đang có nhu cầu tuyển ny giống bn !