\(\frac{7x+2}{5x+7}\)= \(\frac{7x-1}{5x+1}\)
(nhập kết quả dưới dạng phân số tối giản)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}=\frac{\left(7x+2\right)-\left(7x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3}{6}=\frac{1}{2}\)
=> 2(7x + 2) = 5x + 7
14x + 4 = 5x + 7
14x - 5x = 7 - 4
9x = 3
x = 3:9
x = 0,(3)
\(\frac{x+1}{x-1}=\frac{7}{3}\)
=> \(3.\left(x+1\right)=7.\left(x-1\right)\)
=> \(3x+3=7x-7\)
=> \(3x+10=7x\)
=> \(4x=10\)
=> \(x=\frac{10}{4}=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
a)\(\frac{1}{99.97}\)−\(\frac{1}{97.95}\)−\(\frac{1}{95.93}\)−…−\(\frac{1}{5.3}\)−\(\frac{1}{3.1}\)
=\(\frac{1}{99.97}\)−(\(\frac{1}{97.95}\)+\(\frac{1}{95.93}\)+…+\(\frac{1}{5.3}\)+\(\frac{1}{3.1}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(\(\frac{1}{95}\)−\(\frac{1}{97}\)+\(\frac{1}{93}\)−\(\frac{1}{95}\)+…+\(\frac{1}{3}\)−\(\frac{1}{5}\)+1−\(\frac{1}{3}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(1−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).\(\frac{96}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48.99}{99.97}\)
=\(\frac{-4751}{9603}\)
ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)\(\frac{1}{18}\)
= \(\frac{1}{3}-\frac{1}{18}\)
= \(\frac{5}{18}\)
\(x\ne-\frac{7}{5};x\ne-\frac{1}{5}\)
Đề \(\Leftrightarrow\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow35x^2+7x+10x+2=35x^2+49x-5x-7\)
\(\Leftrightarrow17x-44x=-2-7\)
\(\Rightarrow-27x=-9\Rightarrow x=\frac{1}{3}\) (thỏa)
Vậy x = 1/3
Phân số thứ 20 à , hơi khó đó
Nhưng kết quả là:\(\frac{1}{1599}\)