3/35+3/63+3/99+...+3/x(x+2)=24/35
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\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)
\(\frac{3}{10}-\frac{3}{2x+4}=\frac{24}{35}\)
\(\frac{3}{2x+4}=\frac{-27}{70}\)
tự làm nốt
\(\Leftrightarrow\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{5}\)
=>x+2=5
hay x=3
\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)
Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{9.11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
Thay A vào biểu thức
\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{22}{15}\)
P/s: Có thể tính sai :(
\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)
Trước tiên mình tính dãy có dấu ngoặc đã
Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)
Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :
\(\frac{5}{11}\times x=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)
\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)
\(\Leftrightarrow x=\frac{22}{15}\)
Vậy \(x=\frac{22}{15}\)
=>x(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+99/x=-3/7
=>8/9x+99/x=-3/7
\(\Leftrightarrow\dfrac{8x}{9}+\dfrac{99}{x}=\dfrac{-3}{7}\)
\(\Leftrightarrow\dfrac{8x^2+99\cdot9}{9x}=\dfrac{-3}{7}\)
\(\Leftrightarrow-56x^2-6237=27x\)
hay \(x\in\varnothing\)
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
Tk mình đi mọi người mình bị âm nè!
ai tk mình mình tk lại cho!!!
( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 ) x X = 2/3
X = 2/3 : ( 1/13 + 1/15 + 1/35 + 1/63 + 1/99 )
X = 286/85
k mình đi mình đang bị âm
#)Giải :
\(200-18:\left(372:3x-1\right)-28=166\)
\(\Leftrightarrow200-18:\left(372:3x-1\right)=194\)
\(\Leftrightarrow18:\left(372:3x-1\right)=6\)
\(\Leftrightarrow372:3x-1=3\)
\(\Leftrightarrow3x-1=124\)
\(\Leftrightarrow3x=125\)
\(\Leftrightarrow x=\frac{125}{3}\)
200 - 18 : (372 : 3 . x - 1) - 28 = 166
=> 200 - 18 : (372 : 3.x - 1) = 166 + 28
=> 200 - 18 : (372 : 3.x) - 1) = 194
=> 18 : (372 : 3.x - 1) = 200 - 194
=> 18 : (372 : 3.x - 1) = 6
=> 372 : 3.x - 1 = 18 : 6
=> 372 : 3.x - 1 = 3
=> 372 : 3.x = 3 + 1
=> 372 : 3.x = 4
=> 3.x = 372 : 4
=> 3.x = 93
=> x = 93 : 3
=> x = 31
Sửa đề
\(\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{-9}{35}\)
\(x+2=35\)
\(x=35-2\)
\(x=33\)
\(-17x=-17-\left(-34\right)\)
\(-17x=-17\)
\(x=-1\)
Bạn coi lại đề: $\frac{3}{69}$ không nằm trong các số hạng có quy luật.