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Ta có : \(\frac{x}{y}=\frac{2}{3}\Rightarrow x=\frac{2}{3}y\)
Thay \(x=\frac{2}{3}y\)vào A , ta được :
\(A=\frac{5.\frac{2}{3}y+3y}{6.\frac{2}{3}y-7y}\)
\(\Rightarrow A=\frac{\frac{10}{3}y+3y}{4y-7y}\)
\(\Rightarrow A=\frac{\left(\frac{10}{3}+3\right)y}{-3y}\)
\(\Rightarrow A=\frac{\frac{19}{3}y}{-3y}\)
\(\Rightarrow A=\frac{\frac{19}{3}}{-3}\)
\(\Rightarrow A=\frac{19}{3}.-\frac{1}{3}\)
\(\Rightarrow A=-\frac{19}{9}\)
Vậy \(A=-\frac{19}{9}\)
Đặt x=2z;y=3z
=> B=(5x2z+3x3z)/(6x2z-7x3z)
=(19z)/(-9z)
=-19/9
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\Rightarrow\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(=\frac{1+3y+1+7y}{12+4x}=\frac{2+10y}{12+4x}\)
\(=\frac{2\left(1+5y\right)}{2\left(6+2x\right)}=\frac{1+5y}{6+2x}\)
\(\Rightarrow\frac{1+5y}{5x}=\frac{1+5y}{6+2x}\)
\(\Rightarrow5x=6x+2x\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
\(\Rightarrow\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+5y}{5.2}=\frac{1+5y}{10}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{1+5y}{10}\)
\(\Rightarrow10\left(1+3y\right)=12\left(1+5y\right)\)
\(\Rightarrow10+30y=12+60y\)
\(\Rightarrow30y=-2\)
\(\Rightarrow y=-\frac{1}{15}\)
Vậy \(x=2;y=-\frac{1}{15}\)
\(\frac{2x+3y}{x-y}=\frac{2}{3}\)
\(\Leftrightarrow3.\left(2x+3y\right)=2.\left(x-y\right)\)
\(\Leftrightarrow6x+9y=2x-2y\)
\(\Leftrightarrow6x-2x=-2y-9y\)
\(\Leftrightarrow4x=-11y\)
\(\Leftrightarrow\frac{x}{y}=\frac{-11}{4}\)
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
Ta có ;
\(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\end{cases}}\)
Thay x , y vào biểu thức Q , ta có :
\(Q=\frac{5x+3y}{6x-7y}=\frac{5.2k+3.3k}{6.2k-7.3k}=\frac{10k+9k}{12k-21k}=\frac{19k}{-9k}=-\frac{19}{9}\)
=> \(Q=\frac{-19}{9}\)