Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+3y+1-2}{5+7}=\frac{2x+3y-1}{12}\)
\(\Rightarrow\frac{2x+3y-1}{12}=\frac{2x+3y-1}{6x}\)
TH 1 : \(2x+3y-1=0\)
\(\Rightarrow\frac{2x+1}{5}=0;\frac{3y-2}{7}=0\)
\(\Rightarrow2x+1=0;3y-2=0\)
\(\Rightarrow2x=-1;3y=2\)
\(\Rightarrow x=-\frac{1}{2};y=\frac{2}{3}\)
TH 2 : \(2x+3y-1\ne0\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
Mà \(\frac{2x+1}{5}=\frac{3y-2}{7}\)
\(\Rightarrow\frac{2.2+1}{5}=\frac{3y-2}{7}\)
\(\Rightarrow1=\frac{3y-2}{7}\)
\(\Rightarrow3y-2=7\)
\(\Rightarrow3y=9\)
\(\Rightarrow y=3\)
Vậy \(\orbr{\begin{cases}x=-\frac{1}{2};y=\frac{2}{3}\\x=2;y=3\end{cases}}\)
Theo t/c dãy tỉ số bằng nhau :
\(\Rightarrow\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+1+3y-2}{5+7}=\frac{2x+3y-1}{12}\)
Do \(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)
\(\Rightarrow6x=12\Leftrightarrow x=2\)
Xét :\(\frac{2x+1}{5}=\frac{3y-2}{7}\)
\(1=\frac{3y-2}{7}\)
\(\Rightarrow3y=9\Leftrightarrow y=3\)
ta có: \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+1+3y-2}{5+7}=\frac{2x+3y-1}{12}\)
\(\Rightarrow\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)
=> 6x = 12
x = 2
=> \(\frac{2x+1}{5}=\frac{2.2+1}{5}=\frac{5}{5}=1\)
\(\frac{3y-2}{7}=1\Rightarrow3y-2=7\Rightarrow3y=9\Rightarrow y=3\)
KL: x = 2; y = 3
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)
Ta có: \(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)
\(\Rightarrow6x=12\)
\(x=\frac{12}{6}\)
\(x=2\)
\(\Rightarrow\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2.2+1}{5}=1\)
\(\Rightarrow\frac{3y-2}{7}=1\)
\(\Rightarrow3y=7+2\)
\(3y=9\)
\(\Rightarrow y=3\)
Vậy \(x=2;y=3\)
Tham khảo nhé~
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