rút gọn \(M=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{200.201.202.203}\)
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C
3
LN
21 tháng 9 2017
\(S_n=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(S_n=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{n\left(n+2\right)\left(n+3\right)}\right)\)\(S_n=\dfrac{1}{3}\left(\dfrac{1}{2.3.4}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)
\(S_n=\dfrac{1}{3}\left(\dfrac{1}{24}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)
\(S_n=\dfrac{1}{72}-\dfrac{1}{3\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
5 tháng 1 2022
\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)
\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)
26 tháng 9 2021
Ta có \(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\dfrac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
Áp dụng:
\(\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29\cdot30}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+...+\dfrac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{24360}\right)=\dfrac{1}{3}\cdot\dfrac{1353}{8120}=\dfrac{451}{8120}\)
26 tháng 9 2021
\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{27.28.29.30}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)=\dfrac{1}{3}.\dfrac{4060-1}{28.29.30}\)
\(=\dfrac{1}{3}.\dfrac{4059}{24360}=\dfrac{1353}{24360}=\dfrac{451}{8120}\)
NN
1
2 tháng 7 2022
Sửa đề: \(P=\dfrac{2}{2x+3}+\dfrac{3}{2x+1}-\dfrac{6x+5}{\left(2x+1\right)\left(2x+3\right)}\)
\(=\dfrac{4x+2+6x+9-6x-5}{\left(2x+1\right)\left(2x+3\right)}\)
\(=\dfrac{4x+6}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2}{2x+1}\)
BF
1
28 tháng 5 2021
\(D=\dfrac{5}{2x^2+6x}-\dfrac{4-3x^2}{x^2-9}-3\) (đk:\(x\ne3;x\ne-3\))
\(=\dfrac{5}{2x\left(x+3\right)}-\dfrac{4-3x^2}{\left(x-3\right)\left(x+3\right)}-3\)
\(=\dfrac{5\left(x-3\right)}{2x\left(x-3\right)\left(x+3\right)}-\dfrac{\left(4-3x^2\right).2x}{2x\left(x-3\right)\left(x+3\right)}-\dfrac{3.2x\left(x-3\right)\left(x+3\right)}{2x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{5x-15-8x+6x^3-6x\left(x^2-9\right)}{2x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{51x-15}{2x\left(x-3\right)\left(x+3\right)}\)
21 tháng 10 2021
\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
30 tháng 10 2018
Xét mẫu số của phân số:
\(\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}\)
\(=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+\left(\dfrac{99}{1}-98\right)\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Ta thấy mẫu số gấp tử số 100 lần. Vậy phân số đó có giá trị bằng \(\dfrac{1}{100}\)