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a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\frac{1}{2}.B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)
\(B=\frac{100}{101}.2=\frac{200}{101}\)
b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)
\(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
\(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)
\(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)
\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)
\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\cdot\left(1-\frac{1}{101}\right)\)
\(=2\cdot\frac{100}{101}\)
\(=1\frac{99}{101}\)
\(C=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+...+\frac{5}{101.105}\)
\(C=5.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{101.105}\right)\)
\(C=5.\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{101}-\frac{1}{105}\right)\)
\(C=\frac{5}{4}.\left(1-\frac{1}{105}\right)\)
\(C=\frac{5}{4}.\frac{104}{105}\)
\(C=\frac{26}{21}\)
Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được
\(A=\frac{12}{1.5}+\frac{12}{5.9}+\frac{12}{9.13}+.............+\frac{12}{101.105}\)
\(=3.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+............+\frac{4}{101.105}\right)\)
\(=3\left(1-\frac{1}{105}\right)\)
\(=3.\frac{104}{105}=\frac{312}{105}\)
\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)
\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(4A=\frac{x+4-1}{x+4}\)
\(A=\frac{x+3}{\text{4(x+4)}}\)
Bạn tự thay rồi tính nhé
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(A=\left(1-\frac{1}{x+4}\right):4\)
Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)
A = \(\left(1-\frac{1}{16}:4\right)\)
A = \(\frac{15}{16}:4=\frac{15}{64}\)
Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)
A = \(\left(1-\frac{1}{6}\right):4\)
A \(=\frac{5}{6}:4=\frac{5}{24}\)
Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)
A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)
A = \(\frac{23}{29}:4=\frac{23}{116}\)
\(a,x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=--\frac{37}{45}.\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{37}{45}\)
\(x+\frac{1}{5}=\frac{37}{45}+\frac{1}{45}=\frac{38}{45}\)
\(x=\frac{38}{45}-\frac{1}{5}=\frac{29}{45}\)
\(b,\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Rightarrow5x+6=2016\)
\(\Rightarrow5x=2010\Rightarrow x=402\)
\(c,\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}=\frac{2017}{2018}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2017}{2018}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{2017}{2018}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2017}{2018}=\frac{1}{2018}\)
\(\Rightarrow x+2=2018\Rightarrow x=2016\)
học tốt ~~~
1, Tính tổng:
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)
2, Tìm x:
\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)
- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}\cdot-\frac{7}{11}\)
\(=-\frac{5}{11}\)
B=5/1.5+5/5.9+5/9.13+...........+5/101.105
=(4/1.5+4/5.9+4/9.13+..........+4/101.105).5/4
=(1/1-1/5+1/5-1/9+1/9-1/13+..........+1/101-1/105).5/4
=(1/1-1/105).5/4
=104/105.5/4
=26/21
\(B=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+.......+\frac{5}{101.105}\)
\(B=5.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.........+\frac{1}{101}-\frac{1}{105}\right)\)
\(B=5.\left(1-\frac{1}{105}\right)\)
\(B=\frac{5.104}{105}=\frac{104}{21}\)
nha m.n!