Bài 5: (0,5 điểm) Tính giá trị biểu thức \(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
Tính giá trị biểu thức
A=\(\dfrac{1}{2.5}\)+\(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+.....+\(\dfrac{1}{92.95}\)+\(\dfrac{1}{95.98}\)
Mong mn giúp đỡ
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{92}+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(A=\dfrac{49}{98}-\dfrac{1}{98}\)
\(A=\dfrac{48}{98}\)
\(A=\dfrac{24}{49}\)
Giải thích các bước giải:
A =1/2.5 + 1/5.8 + 1/8.11 + … +1/92.95 + 1/95.98
=1/3 . (1/2-1/5+1/5-1/8+1/8-1/11+…+1/92-1/95+1/95-1/98)
=1/3 . (1/2 – 1/98 )
=1/3 . 24/49
=8/49`
vậy `A=8/49`
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)
\(3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(=\dfrac{1}{2}-\dfrac{1}{98}\)
\(=\dfrac{24}{49}\)
\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{96}{196}=\dfrac{8}{49}\)
tính biểu thức : A = \(\dfrac{1}{2.5}\)+ \(\dfrac{1}{5.8}\)+ \(\dfrac{1}{8.11}\)+...... + \(\dfrac{1}{2012.2015}\)+ \(\dfrac{1}{2015.2018}\)+\(\dfrac{1}{2018.2021}\)
3A=3/2.5+...+3/2018.2021
3A=1/2-1/5+1/5-...+1/2018-1/2021
3A=1/2-1/2021 sau tự tính A
3A= 1/2- 1/5 + 1/5- 1/8+ 1/8 -1/11+...+ 1/2012- 1/2015 +1/2015- 1/2018-1/2021
3A =1/2 -1/2021
3A = 2019/ 4042
=> 2019/4042 : 3 = 673/4042
Chúc bạn học tốt !!
Giải:
A=1/2.5+1/5.8+1/8.11+...+1/2012.2015+1/2015.2018+1/2018.2021
A=1/3.(3/2.5+3/5.8+3/8.11+...+3/2012.2015+3/2015.2018+3/2018.2021)
A=1/3.(1/2-1/5+1/5+1/8+1/8-1/11+...+1/2012-1/2015+1/2015-1/2018+1/2018-1/2021)
A=1/3.(1/2-1/2021)
A=1/3.2019/4042
A=673/4042
Chúc bạn học tốt!
A= 1/2.5+1/5.8+1/8.11+...+1/92.95+1/95.98.
Tính tổng A
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{95}-\frac{1}{98}\)
\(=\frac{1}{2}-\frac{1}{98}\)tự làm tiếp
\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)\(\dfrac{3}{11.14}\)+\(\dfrac{3}{14.17}\)<\(\dfrac{1}{2}\)
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
A=1/2.5+1/5.8+1/8.11+...+1/92.95+1/95.98
3A = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\)
3A=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)
3A=\(\frac{1}{2}-\frac{1}{98}\)
3A=\(\frac{98}{196}-\frac{2}{196}\)=\(\frac{96}{196}=\frac{24}{49}\)
A=\(\frac{24}{49}:3=\frac{24}{49}.\frac{1}{3}=\frac{8}{49}\)
Vậy A = \(\frac{8}{49}\)
\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)
\(\Rightarrow3A=3\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\right)\)
\(\Rightarrow3A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\)
\(\Rightarrow3A=\frac{1}{2}-\frac{1}{98}\)
\(\Rightarrow3A=\frac{24}{49}\)
\(\Rightarrow A=\frac{24}{49}:3\)
\(\Rightarrow A=\frac{8}{49}\)
Vậy \(A=\frac{8}{49}\)
\(A=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(=3.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(=3.\frac{24}{49}\)
\(=\frac{72}{49}\)
A= 1/2.5+1/5.8+1/8.11+...+1/92.95+1/95.98.
Tính tổng A?
A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98
A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )
A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )
A = 1/3 . ( 1/2 - 1/98 )
A = 1/3 . 24/49
A = 8/49
A = 1/2x5 + 1/5x8 + 1/8x11 + ... + 1/92x95 + 1/95x98
A = 1/3 x ( 3/2x5 + 3/5x8 + 3/8x11 + ... + 3/92x95 + 3/95x98 )
A = 1/3 x ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/95 - 1/98 )
A = 1/3 x ( 1/2 - 1/98 )
A = 1/3 x 24/29
A = 8/49
A = 1/2x5 + 1/5x8 + 1/8x11 + ... + 1/92x95 + 1/95x98
A = 1/3 x ( 3/2x5 + 3/5x8 + 3/8x11 + ... + 3/92x95 + 3/95x98 )
A = 1/3 x ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/95 - 1/98 )
A = 1/3 x ( 1/2 - 1/98 )
A = 1/3 x 24/29
A = 8/49
Tìm x biết: \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+\dfrac{3x}{8.11}+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{3}{2.5}\) \(+
\) \(\dfrac{3}{5.8}\) \(+\) \(\dfrac{3}{8.11}\) \(+\) \(\dfrac{3}{11.14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{8}\) \(+\) \(\dfrac{1}{8}\) \(-\) \(\dfrac{1}{11}\) \(+\) \(\dfrac{1}{11}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) x \(\dfrac{3}{7}\) \(=\) \(\dfrac{1}{21}\)
\(x\) \(=\) \(\dfrac{1}{21}\) \(:\) \(\dfrac{3}{7}\)
\(x\) \(=\) \(\dfrac{1}{9}\)