$\frac{0,2}{|2x-1|}$ -$\frac{20}{11.13}$ -$\frac{20}{13.15}$ -…-$\frac{20}{53.55}$ = $\frac{3}{11}$ 05/08/2021 Bởi aihong $\frac{0,2}{|2x-1|}$ -$\frac{20}{11.13}$ -$\frac{20}{13.15}$ -…-$\frac{20}{53.55}$ = $\frac{3}{11}$
Đáp án + Giải thích các bước giải: `(0,2)/|2x-1| – 20/(11*13)-20/(13*15)-…-20/(53*55)=3/11` $\\$ `=> (0,2)/|2x-1|-(20/(11*13)+20/(13*15)+…+20/(53*55))=3/11` $\\$ `=> (0,2)/|2x – 1| – 20/2(2/(11*13)+2/(13*15)+…+2/(53*55))=3/11` $\\$ `=> (0,2)/|2x-1| -10(1/11 – 1/13 + … + 1/53 – 1/55) = 3/11` $\\$ `=> (0,2)/|2x – 1| – 10(1/11 – 1/55) = 3/11` $\\$ `=> (0,2)/|2x-1| – 8/11 = 3/11` $\\$ `=> (0,2)/|2x – 1| = 1` $\\$ `=> |2x – 1| = 0,2` $\\$ `=> `\(\left[ \begin{array}{l}2x-1=0,2\\2x-1=-0,2\end{array} \right.\) $\\$ `=> `\(\left[ \begin{array}{l}x=0,6\\x=0,4\end{array} \right.\) Vậy `x in {0,6;0,4}` Bình luận
`(0,2)/(|2x-1|) – 20/11.13 – 20/13.15 – …- 20/53.55= 3/11` `(0,2)/(|2x-1|) – (20/11.13 + 20/13.15 +…+ 20/53.55)= 3/11` Đặt `A= 20/11.13 + 20/13.15 +…+ 20/53.55` `A=10 ( 2/11.13 + 2/13.15 + ….+ 2/53.55)` `A= 10 ( 1/11 – 1/13 + 1/13 – 1/15 +…+ 1/53 – 1/55)` `A= 10 ( 1/11 – 1/55)` `A= 10. 4/55` `A= 8/11` `=> (0,2)/(|2x-1|) – 8/11 = 3/11` `=> (0,2)/(|2x-1|) = 3/11 + 8/11` `=> (0,2)/(|2x-1|) = 1` `=> |2x-1| = 0,2` `=>` \(\left[ \begin{array}{l}2x-1=0,2\\2x-1=-0,2\end{array} \right.\) `=>` \(\left[ \begin{array}{l}2x= 1,2\\2x=0,8\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=0,6\\x=0,4\end{array} \right.\) Vậy `x= 0,6` hoặc `x= 0,4` Bình luận
Đáp án + Giải thích các bước giải:
`(0,2)/|2x-1| – 20/(11*13)-20/(13*15)-…-20/(53*55)=3/11` $\\$ `=> (0,2)/|2x-1|-(20/(11*13)+20/(13*15)+…+20/(53*55))=3/11` $\\$ `=> (0,2)/|2x – 1| – 20/2(2/(11*13)+2/(13*15)+…+2/(53*55))=3/11` $\\$ `=> (0,2)/|2x-1| -10(1/11 – 1/13 + … + 1/53 – 1/55) = 3/11` $\\$ `=> (0,2)/|2x – 1| – 10(1/11 – 1/55) = 3/11` $\\$ `=> (0,2)/|2x-1| – 8/11 = 3/11` $\\$ `=> (0,2)/|2x – 1| = 1` $\\$ `=> |2x – 1| = 0,2` $\\$ `=> `\(\left[ \begin{array}{l}2x-1=0,2\\2x-1=-0,2\end{array} \right.\) $\\$ `=> `\(\left[ \begin{array}{l}x=0,6\\x=0,4\end{array} \right.\)
Vậy `x in {0,6;0,4}`
`(0,2)/(|2x-1|) – 20/11.13 – 20/13.15 – …- 20/53.55= 3/11`
`(0,2)/(|2x-1|) – (20/11.13 + 20/13.15 +…+ 20/53.55)= 3/11`
Đặt `A= 20/11.13 + 20/13.15 +…+ 20/53.55`
`A=10 ( 2/11.13 + 2/13.15 + ….+ 2/53.55)`
`A= 10 ( 1/11 – 1/13 + 1/13 – 1/15 +…+ 1/53 – 1/55)`
`A= 10 ( 1/11 – 1/55)`
`A= 10. 4/55`
`A= 8/11`
`=> (0,2)/(|2x-1|) – 8/11 = 3/11`
`=> (0,2)/(|2x-1|) = 3/11 + 8/11`
`=> (0,2)/(|2x-1|) = 1`
`=> |2x-1| = 0,2`
`=>` \(\left[ \begin{array}{l}2x-1=0,2\\2x-1=-0,2\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x= 1,2\\2x=0,8\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0,6\\x=0,4\end{array} \right.\)
Vậy `x= 0,6` hoặc `x= 0,4`