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A=5/2x(2/2x4+2/4x6+2/6x8+...+2/14x16)
=5/2x(1/2-1/4+1/4-1/6+...+1/14-1/16)
=5/2x(1/2-1/16)
=5/2x(7/16)
=35/32
Giải
1/2x4+1/4x6+1/6x8+...+1/96x98+1/98x100
= 1/2 x (1/2 - 1/4 + 1/4 - 1/6 + 1/6-1/8 + ... + 1/98 - 1/100)
= 1/2 x (1/2 - 1/100)
= 1/2 x 98/100
= 98/200
ĐS: 98/200
Để giải phương trình \( y:(\frac{1}{2} \times 4+\frac{1}{4} \times 6+\frac{1}{6} \times 8+\frac{1}{8} \times 10) \times y=\frac{1}{3} \), ta có thể làm như sau:
Đầu tiên, tính giá trị của phần tử ngoặc đơn trong phương trình:
\( \frac{1}{2} \times 4+\frac{1}{4} \times 6+\frac{1}{6} \times 8+\frac{1}{8} \times 10 \).
\( = \frac{2}{2} \times 4+\frac{1}{2} \times 6+\frac{1}{3} \times 8+\frac{1}{4} \times 10 \).
\( = 2+3+\frac{8}{3}+\frac{10}{4} \).
\( = 2+3+\frac{8}{3}+2.5 \).
\( = 5+2.667+2.5 \).
\( = 10.167 \).
Tiếp theo, thay giá trị tính được vào phương trình:
\( y \times 10.167 = \frac{1}{3} \).
Để tìm giá trị của y, ta chia cả hai vế của phương trình cho 10.167:
\( y = \frac{\frac{1}{3}}{10.167} \).
Tiếp tục tính toán:
\( y = \frac{1}{3} \times \frac{1}{10.167} \).
\( y \approx 0.030 \).
Vậy giá trị của y là khoảng 0.030.
[1/(2 × 4) + 1/(4 × 6) + 1/(6 × 8) + 1/(8 × 10)] × y = 1/3
(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + 1/8 - 1/10) × y = 1/3
(1/2 - 1/10) × y = 1/3
2/5 × y = 1/3
y = 1/3 : 2/5
y = 5/6
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
Xx0,5+Xx0,25x6+Xx0,5x4=8
Xx0,5+Xx1,5+Xx2=8
Xx(0,5+1,5+2)=8
Xx4 =8
X =8:4
X =2
A=\(\frac{2}{2x4}+\frac{2}{4x6}+.........+\frac{2}{2014x2016}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.............+\frac{1}{2014}-\frac{1}{2016}\)
=\(\frac{1}{2}-\frac{1}{2016}\)
=\(\frac{1008}{2016}-\frac{1}{2016}\)
=\(\frac{1007}{2016}\)
2[1/2X4+1/4X6+1/6X8+...+1/Xx(X+2)]=11/45x2
2/2x4+2/4x6+2/6x8+....+2/Xx(X+2)=22/45
1/2-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2=22/45
1/2-1/x+2=22/45
1/x+2=1/2-22/45
1/x+2=1/90
=>x+2=90
=>x=88
vậy x=88
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{45}\)
\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{22}{45}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{22}{45}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{90}\)
=>x+2=90
=>x=90-2
=>x=88
vậy x=88
x:2 +x:4x6 +x :2x4 = ?
=> x:2+x:24+x:8 = ?
=> x:(2+24+8) =?
=> x:34 = ?
(đề bài thiếu hả bạn)
`(2/(2xx4)+2/(4xx6)+2/(6xx8)+2/(8xx10))xxy=1/3`
`=>(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10)xxy=1/3`
`=>(1/2-1/10)xxy=1/3`
`=>(5/10-1/10)xxy=1/3`
`=>4/10xxy=1/3`
`=>2/5xxy=1/3`
`=>y=1/3:2/5`
`=>y=1/3xx5/2`
`=>y=5/6`
Đặt A = \(\dfrac{5}{2\cdot4}+\dfrac{5}{4\cdot6}+...+\dfrac{5}{132\cdot134}\)
\(\dfrac{2}{5}A=\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{132\cdot134}\)
\(\dfrac{2}{5}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{132}-\dfrac{1}{134}\)
\(\dfrac{2}{5}A=1-\dfrac{1}{134}=\dfrac{133}{134}\)
A = \(\dfrac{133}{134}:\dfrac{2}{5}=\dfrac{665}{268}\)
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+...+\dfrac{5}{132.134}\)
\(=2,5\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{132.134}\right)\)
\(=2,5\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{134-132}{132.134}\right)\)
\(=2,5\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...-\dfrac{1}{134}\right)\)
\(=2,5\left(\dfrac{1}{2}-\dfrac{1}{134}\right)\)
\(=2,5.\dfrac{33}{67}\)
\(=\dfrac{165}{134}\)