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Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)
Vậy A < 2
Bài 3:
D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)
\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)
Bài 4:
A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)
\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
Ta có :
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)\)
\(=100-1-\dfrac{1}{2}-\dfrac{1}{3}-..................-\dfrac{1}{100}\)
\(=99-\dfrac{1}{2}-\dfrac{1}{3}-................-\dfrac{1}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+..................+\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+.................+\dfrac{99}{100}\)
Vậy :\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+....................+\dfrac{99}{100}\)
\(\Rightarrowđpcm\)
1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3
=>x=-16/3:7/3=-7/16
2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35
=>|x-2|=129/35
=>x-2=129/35 hoặc x-2=-129/35
=>x=199/35 hoặc x=-59/35
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Ta có:\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)\(=\left(1+1+...+1\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=100-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)(đpcm)
Bài 1: Tìm x biết:
a) \(\dfrac{6}{5}-2\left|1-3x\right|=1\dfrac{2}{3}\)
\(2\left|1-3x\right|=\dfrac{6}{5}-1\dfrac{2}{3}\)
\(2\left|1-3x\right|=\dfrac{-7}{15}\)
\(\left|1-3x\right|=\dfrac{-7}{15}:2\)
\(\left|1-3x\right|=\dfrac{-7}{30}\)
\(\left|1-3x\right|\in N\) nhưng \(\dfrac{-7}{30}\notin N\)
\(\Rightarrow x=\varnothing\)
b) \(\left(2,8x+50\right):\dfrac{-3}{2}=51\)
\(\left(2,8x+50\right)=51.\dfrac{-3}{2}\)
\(2,8x+50=\dfrac{-153}{2}\)
\(2,8x=\dfrac{-153}{2}-50\)
\(2,8x=\dfrac{-253}{2}\)
\(x=\dfrac{-253}{2}:2,8\)
\(x=\dfrac{-1265}{28}\)
c) \(\dfrac{x-2}{-2}=\dfrac{x+4}{3}\)
\(\Rightarrow\left(x-2\right).3=-2.\left(x+4\right)\)
\(x.3-2.3=\left(-2\right).x+\left(-2\right).4\)
\(3x-6=\left(-2\right)x+\left(-8\right)\)
\(3x-\left(-2\right)x=6+\left(-8\right)\)
\(5x=-2\)
\(x=\left(-2\right):5\)
\(x=\dfrac{-2}{5}\)
d) \(4\left(3-2x\right)-5\left(x-1\right)=12\)
\(4.3-4.2x-5x+5.1=12\)
\(12-8x-5x+5=12\)
\(12+\left(-8\right)x+\left(-5\right)x+5=12\)
\(12+\left(-13\right)x+5=12\)
\(\left(-13\right)x=12-12-5\)
\(\left(-13\right)x=-5\)
\(x=\left(-5\right):\left(-13\right)\)
\(x=\dfrac{5}{13}\)
Bài 2: Chứng minh:
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\) (đpcm)
2)
\(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3}{3}+\dfrac{1}{3}+\dfrac{3^2}{3^2}+\dfrac{1}{3^2}+\dfrac{3^3}{3^3}+\dfrac{1}{3^3}+...+\dfrac{3^{98}}{3^{98}}+\dfrac{1}{3^{98}}\\ D=1+\dfrac{1}{3}+1+\dfrac{1}{3^2}+1+\dfrac{1}{3^3}+...+1+\dfrac{1}{3^{98}}\\ D=\left(1+1+1+...+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ D=98+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\)
Gọi \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\) là \(C\)
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ 3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ 2C=1-\dfrac{1}{3^{98}}\\ C=\left(1-\dfrac{1}{3^{98}}\right):2\\ C=1:2-\dfrac{1}{3^{98}}:2\\ C=\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}\)
\(D=98+C=98+\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}=98\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}< 100\)
Vậy \(D< 100\)