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Gọi tử số của phân số cần tìm là x
Ta có: \(\dfrac{-3}{2}\) = \(\dfrac{x}{-22}\)
=> (-3) . (-22) : 2 = x
=> x = 33
Vậy phân số đó là \(\dfrac{33}{-22}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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C= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\) (Đặt lần lượt 2 và 3 ở tử và mẫu ra ngoài)
= \(\frac{2}{3}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
\(C=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(A=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
\(A=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}\)
\(=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(\frac{x+\frac{2}{7}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}-\frac{2}{3}=0\)
=>\(\frac{x+0,396535406}{1,194803109}-\frac{2}{3}=0\)
=>x+0,396535406:1,194803109-2/3=0
x+0,396535406:1,194803109=0+2/3=2/3
còn lại tự giải nha nhiều quá
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F = \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}-\frac{3}{203}+\frac{3}{17}+\frac{3}{5}}=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
F = \(\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x. (22+23)=90
x.45=90
x=90:45
x=2
b)2x+293=309
2x=309-293
2x=16
x=16:2
x=8
ta có: \(A=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow2A=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(\Rightarrow2A-A=6-\frac{3}{2^9}\)
\(A=6-\frac{3}{2^9}\)