Lm câu 2 trc nhé:
\(x-3+x-3=\left(x-3\right)+\left(x-3\right)=2\left(x-3\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
Chỉ lm tắt thôi ạ, hiểu rồi tự trình bày nha~
\(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x}{3}-\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x5-3}{15}\)
\(\Leftrightarrow4.15=x5-3y\)
\(\Leftrightarrow60=x5-3y\)
\(\Leftrightarrow x5-3y=60\)
tìm x,y như bt nhé
\(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\Rightarrow\frac{4}{y}=\frac{1}{5}+\frac{x}{3}\Rightarrow\frac{4}{y}=\frac{3}{15}+\frac{5x}{15}\Rightarrow\frac{4}{y}=\)\(\frac{3+5x}{15}\)
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đến đây tự làm nhé
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hay để t lm cho
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ko mún làm nữa
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lần sau nhé
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pái pái
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1.\(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
\(\Rightarrow\frac{4}{y}=\frac{x}{3}-\frac{1}{5}\)
\(\Rightarrow\frac{4}{y}=\frac{5x-3}{15}\)
\(\Rightarrow4.15=y.\left(5x-3\right)\)
\(\Rightarrow y.\left(5x-3\right)=60\)
mà \(x;y\in N\)
\(\Rightarrow y.\left(5x-3\right)=1.60=2.30=3.20=4.15=5.12=6.10\)
Ta có bảng sau:
| \(5x-3\) | \(1\) | \(60\) | \(2\) | \(30\) | \(3\) | \(20\) | \(4\) | \(15\) | \(5\) | \(12\) | \(6\) | \(10\) |
| \(y\) | \(60\) | \(1\) | \(30\) | \(2\) | \(20\) | \(3\) | \(15\) | \(4\) | \(12\) | \(5\) | \(10\) | \(6\) |
| \(x\) | \(\frac{4}{5}\) | \(\frac{63}{5}\) | \(1\) | \(\frac{33}{5}\) | \(\frac{6}{5}\) | \(\frac{23}{5}\) | \(\frac{7}{5}\) | \(\frac{18}{5}\) | \(\frac{8}{5}\) | \(3\) | \(\frac{9}{5}\) | \(\frac{13}{5}\) |
Vậy \(\left(x;y\right)\in\left\{\left(1;30\right);\left(3;5\right)\right\}\)
2.\(x-3+x-3=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
