15 x {X+35}+35=575
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= 2/3×5 + 2/ 5×7 + 2/ 7×9 + 2/ 9×11 + ...... + 2/ 23×25+ 2 / 25×27
=1/3 - 1/5 + 1/5 -1/7 + 1/7 -1/9 +1/9 -1/11+.......+1/23- 1/25 + 1/25 -1/27
= 1/3 - ( 1/5+1/5 -1/7 +1/7 - 1/9 +1/9-1/11 +........+ 1/23 -1/25 + 1/25 )-1/27
=1/3 - 1/27= 9/27 - 1/27= 8/27
1/1x3 + 1/3x5 + 1/5x7 + 1/7x9 +... + 1/23x25 + 1/25x27 =
1/2 x (2/1x3 + 2/3x5 + 2/5x7 +..... 2/23x25 +2/25x27 =
1/2 x (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +...... + 1/23 - 1/25 + 1/25 - 1/27 =
1/2 x (1 - 1/27) =
1/2 x 26/27 = 13/27
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{575}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{23.25}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{3}-\frac{1}{25}=\frac{22}{75}\)
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)
=> \(A=\frac{121}{243}\)
=> \(2x+\frac{12}{25}=\frac{121}{243}\)
=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
=> x = ......
Bài 6. Tìm x ϵ N biết
a) (x –15) .15 = 0
b) 32 (x –10 ) = 32
c) ( x – 5)(x – 7) = 0
d) (x – 35).35 = 35
A.\(\left(x-15\right).15=0\)
\(x-15=0:15\)
\(x-15=0\)
\(x=15+0\)
\(x=15\)
B.\(32\left(x-10\right)=32\)
\(x-10=32:32\)
\(x-10=1\)
\(x=10+1\)
\(x=11\)
`a) `
`(x-15)xx15=0`
`<=> x-15 = 0 : 15`
`<=> x-15 = 0`
`<=> x = 0 + 15`
`<=> x =15`
`b)`
`32.(x-10)=32`
`<=> x - 10 = 32:32`
`<=>x-10=1`
`<=> x = 1+10`
`<=> x =11`
`c)`
`(x-5).(x-7)=0`
`<=>` \(\left[ \begin{array}{l}x-5 = 0\\x-7=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
`d)`
`(x-35)xx35=35`
`<=> x - 35 = 35:35`
`<=> x - 35 = 1`
`<=> x = 1+35`
`<=> x = 36`
\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{575}\)
\(\Rightarrow A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{23.25}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{25}\)
\(\Rightarrow A=\frac{25}{75}-\frac{3}{75}\)
\(\Rightarrow A=\frac{22}{75}\)
\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{575}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{23.25}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}\)
\(=\frac{1}{3}-\frac{1}{25}\)
\(=\frac{22}{75}\)
Study well ! >_<
15 x {X+35}+35=575
15 x {X+35}=575-35
15 x {X+35}=540
{X+35}=540:15
{X+35}=36
X=36-35
X=1