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\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
Nhớ k cho m nhé!
b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
D = \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)
D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)
Chắc vậy
\(3\times\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)
=> \(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}=\frac{20}{13}\)
=> \(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}=\frac{20}{13}\)
=> \(\frac{x}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{20}{13}\)
=> \(\frac{x}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{13}\right)=\frac{20}{13}\)
=> \(\frac{x}{3}\left(1-\frac{1}{13}\right)=\frac{20}{13}\)
=> \(\frac{x}{3}\cdot\frac{12}{13}=\frac{20}{13}\)
=> \(\frac{x}{3}=\frac{20}{13}:\frac{12}{13}=\frac{20}{13}\cdot\frac{13}{12}=\frac{5}{3}\)
=> x = 5
\(3\cdot\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)
\(3\cdot\left(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}\right)=\frac{60}{13}\)
\(3\left(x-3\right)\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{60}{13}\)
\(\left(3x-9\right)\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)=\frac{60}{13}\)
\(\left(3x-9\right)\left(1-\frac{1}{13}\right)=\frac{60}{13}\)
\(\left(3x-9\right)\cdot\frac{12}{13}=\frac{60}{13}\)
\(3x-9=\frac{\frac{60}{13}}{\frac{12}{13}}\)
\(3x-9=5\)
\(3x=5+9\)
\(3x=14\)
\(x=\frac{14}{3}\approx4,667\)
CÁCH LÀM NHƯ SAU :
(7/28 + 1/28) + 1/70 + 1/130 + 1/x.(x+3)
8/28 + 1/70 +1/130 +1/x.(x+3)
2/7+1/70+1/130+1/x.(x+3)
(20/70 +1/70)+1/130+1/x.(x+3)
3/10+1/130+1/x.(x+3)
39/130+1/130+1/x.(x+3)
4/13+1/x.(x+3)
Đến đây bn tự làm hộ mình vớ. chúc hok tốt k cho mình nhé
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{12}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}.\frac{12}{13}+\frac{1}{3}.\frac{1}{x}-\frac{1}{3}.\frac{1}{x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3}.\frac{1}{x+3}\)
\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3x}\)
\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3}.\frac{1}{x}\)
\(=\frac{4}{13}=\frac{1}{3}\left(\frac{1}{x+3}-\frac{1}{x}\right)\)
\(=\frac{4}{13}:\frac{1}{3}=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{12}{13}=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{12}{13}=\frac{x-\left(x+1\right)}{\left(x+1\right)x}\)
\(=\frac{12}{13}=-\frac{1}{x^2+x}\)
\(\Leftrightarrow=12\left(x^2+x\right)=13.\left(-1\right)\)
\(=12\left(x^2+x\right)=-13\)
\(=x^2+x=-\frac{13}{12}\)
\(=x\left(x+1\right)=-\frac{13}{12}\)
.... Chiụ
\(\frac{3}{34}+\frac{34}{34}.\frac{3}{4}=\frac{3}{34}+1.\frac{3}{4}=\frac{3}{34}+\frac{3}{4}=\frac{57}{68}\)
\(\frac{23}{3}.\frac{56}{6}+\frac{86}{78}=\frac{23}{3}.\frac{28}{3}+\frac{43}{39}=\frac{644}{9}+\frac{28}{3}=\frac{728}{9}\)
\(\frac{3}{45}:\frac{1}{4}=\frac{1}{15}.4=\frac{4}{15}\)
\(\frac{5}{34}-\frac{3}{6}=\frac{5}{34}-\frac{1}{2}=\frac{3}{4}\)
Ta có:
\(M=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...\)
Dựa vào quy luật trên=>Số hạng thứ 30 là:\(\frac{1}{98.101}\)
\(\Rightarrow3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{98}-\frac{1}{101}\)
\(\Rightarrow M=\left(1-\frac{1}{101}\right):3=\frac{100}{101}.\frac{1}{3}=\frac{100}{303}\)
Mình viết hơi tắt mong bạn thông cảm.
\(\frac34+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\ldots+\frac{3}{y}=\frac{33}{34}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\cdots+\frac{3}{y}=\frac{33}{44}\)
\(\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{13-10}{10.13}+\cdots+\frac{3}{y}=\frac{33}{34}\)
\(1-\frac14+\frac14-\frac17+\frac17-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\cdots+\frac{3}{y}=\frac{33}{34}\)
\(1+\frac{3}{y}=\frac{33}{34}\)
\(\frac{3}{y}=\frac{33}{34}-1\)
\(\frac{3}{y}=\frac{-1}{34}\) \(\Rightarrow y=\frac{3.34}{-1}\)
\(y=-102\)
Vậy \(y=-102\)
`3/4 + 3/28 + 3/70 + 3/130 + ... +3/y = 33/34`
`=> 3/(1*4) + 3/(4*7) + 3/(7*10) + 3/(10*13) + ... + 3/y = 33/34`
`=> 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + .... + 3/y = 33/34`
`=> 1 + 3/y = 33/34`
`=> 3/y = 33/34 - 1`
`=> 3/y = 33/34 - 34/34`
`=> 3/y = -1/34`
`=> 3 : y = -1/34`
`=> y = 3: -1/34`
`=> y = 3 xx -34`
`=> y = -102`
Vậy `y = -102`