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A= 2( 1/15 + 1/35 + 1/63+ 1/99+1/143)
A= 2(1/3x5 +1/5x7 + 1/7x9 + 1/9x11 + 1/11x13)
A=2(1/3-1/5+1/5-1/+1/7-1/9+1/9-1/11+1/11-1/13)
A=2(1/3-1/13)
A=2x10/39
A=20/39
A = 2/15 + 2/35 + 2/63 + 2/99 + 2/143
A = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13
A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13
A = 1/3 - 1/13
A = 12/13
Giải:
\(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)
\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{1}-\dfrac{1}{13}\)
\(=\dfrac{12}{13}\)
Chúc em học tốt!
2/3+2/15+2/35+2/63+2/99+2/143
=2(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)
=2(1-1/3+1/3-1/5+1/5-....+1/13)
=2(1-1/13)
=2.12/13=24/13
Đặt \(B=\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}\)
\(\Leftrightarrow B=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)
\(\Leftrightarrow2B=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Leftrightarrow2B=3\left(\frac{1}{3}-\frac{1}{13}\right)=1-\frac{3}{13}=\frac{10}{13}\)
\(\Leftrightarrow A=1+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+\frac{3}{143}=1+\frac{10}{13}=\frac{23}{13}\)
\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\frac{10}{11}.y=\frac{4}{3}\)
\(\Rightarrow y=\frac{22}{15}\)
\(C=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)
\(C=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{13}\)
\(C=1-\frac{1}{13}\)
\(C=\frac{12}{13}\)
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(C=\frac{1}{1}-\frac{1}{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(C=\frac{1}{1}-\frac{1}{13}\)
\(C=\frac{12}{13}\)
B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301
B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301
\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);
\(\frac{2}{15}=\frac{1}{3}-\frac{1}{5}\);
\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).
=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)
=> B=\(\frac{1}{1}-\frac{1}{301}\)
=> B=\(\frac{300}{301}\)
mi tích tau tau tích mi xong tau trả lời nka việt nam nói là làm
\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)
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 \(x=\frac{22}{15}\)
#)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
ê bạn ơi tìm y sao lại có x
\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5