Tìm x\(\in\)N biết:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\)
Giải chi tiết nha
Tính kiểu j bn, dãy phân số kia cộng lại bằng bao nhiêu, đầu bài ko cho à??
A= 3/2.5+3/5.8+3/8.11+3/11.14
B=3/2.5+3/5.8+3/8.11+3/11.14
=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14
=1/2-1/14
=7/14-1/14=6/14=3/7
Tìm x biết:
\(3x-\dfrac{15}{5.8}-\dfrac{15}{8.11}-\dfrac{15}{11.14}-...-\dfrac{15}{47.50}=2\dfrac{1}{10}\)
Giúp mik với, giải chi tiết dùm nhe :>
`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`
`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`
`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`
`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`
`3x-5(1/5-1/50)=21/10`
`3x-5*9/50=21/10`
`3x-9/10=21/10`
`3x=21/10+9/10`
`3x=3`
`x=1`
Bài 1:Tính A=3/5.8+3/8.11+3/11.14+...+3/202.205.
. là dấu nhân nha!
3/5.8 + 3/8.11 + 3/11.14 + ...... + 3/105.108
Chú thích*:
/=phần
.=nhân
giải hộ mình với
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(=\frac{1}{5}-\frac{1}{108}\)
\(=\frac{108}{540}-\frac{5}{540}=\frac{103}{540}\)
Ta có:
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
= \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
= \(\frac{1}{5}-\frac{1}{108}\)
= \(\frac{103}{540}\)
#)Giải :
Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(A=\frac{1}{5}-\frac{1}{108}\)
\(A=\frac{103}{540}\)
\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)\(\dfrac{3}{11.14}\)+\(\dfrac{3}{14.17}\)<\(\dfrac{1}{2}\)
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
Tính : \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2009.2012}+\frac{3}{2012.2015}\)
Ta có :
\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2009.2012}+\frac{3}{2012.2015}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2009}-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2015}\)
\(A=\frac{1}{5}-\frac{1}{2015}\)
\(A=\frac{402}{2015}\)
Vậy \(A=\frac{402}{2015}\)
Chúc bạn học tốt ~
\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2012.1015}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\)
\(A=\frac{1}{5}-\frac{1}{2015}\)
\(A=\frac{402}{2015}\)
3.x/2.5+3.x/5.8+3.x/8.11+3.x/11.14=1/21
Tìm x biết:
3.x/2.5+3.x/5.8+3.x/8.11+3.x/11.14=1/21
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