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a) \(\frac{x+7}{2x+3}-\frac{5}{2x+3}=\frac{x+7-5}{2x+3}=\frac{x+2}{2x+3}\)
b) \(\frac{m^2}{3\left(m+3\right)}+\frac{2m+3}{m+3}=\frac{m^2}{3\left(m+3\right)}+\frac{\left(2m+3\right).3}{3.\left(m+3\right)}\)
\(=\frac{m^2+6m+9}{3\left(m+3\right)}=\frac{\left(m+3\right)^2}{3\left(m+3\right)}=\frac{m+3}{3}\)
c) \(\frac{x^2-4}{x+5}.\frac{2x+10}{x+2}=\frac{\left(x-2\right)\left(x+2\right).2.\left(x+5\right)}{\left(x+5\right).\left(x+2\right)}=\left(x-2\right).2=2x-4\)
d) \(\frac{3+6y}{y^2-2y+1}:\frac{2y+1}{y-1}=\frac{3\left(2y+1\right)}{\left(y-1\right)^2}.\frac{y-1}{2y+1}=\frac{3}{y-1}\)
\(a,\frac{x+7}{2x+3}-\frac{5}{2x+3}\)
\(=\frac{x+7-5}{2x+3}\)
\(=\frac{x+2}{2x+3}\)
\(b,\frac{m^2}{3\left(m+3\right)}+\frac{2m+3}{m+3}\)
\(=\frac{m^2}{3\left(m+3\right)}+\frac{3\left(2m+3\right)}{3\left(m+3\right)}\)
\(=\frac{m^2+6m+9}{3\left(m+3\right)}\)
\(=\frac{\left(m+3\right)^2}{3\left(m+3\right)}\)
\(=\frac{m+3}{3}\)
\(c,\frac{x^2-4}{x+5}.\frac{2x+10}{x+2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{x+5}.\frac{2\left(x+5\right)}{x+2}\)
\(=2\left(x-2\right)\)
d, nghịch đảo lên rồi làm tương tự nha
Do 13;3 là 2 số nguyên tố
=> \(\hept{\begin{cases}m^2+mn+n^2=13k\left(1\right)\\m+2n=3k\end{cases}}\left(k\in Z\right)\)
(1) <=> \(4m^2+4mn+4n^2=52k\)
=> \(3m^2+\left(m+2n\right)^2=52k\)
=> \(3m^2+9k^2=52k\)
=> \(3m^2=k\left(52-9k\right)\)
Do \(m^2\ge0\)
=> \(0< k\le\frac{52}{9}\)
=> \(k\in\left\{1;2;3;4;5\right\}\)
+ \(k=1\)=> \(m=\sqrt{\frac{43}{3}}\left(l\right)\)
+ \(k=2\)=> \(m=\sqrt{\frac{68}{3}}\left(l\right)\)
+ \(k=3\)=> \(\orbr{\begin{cases}m=5\rightarrow n=2\\m=-5\rightarrow n=7\end{cases}}\)
+ \(k=4\)=> \(m=\sqrt{\frac{64}{3}}\left(l\right)\)
+ \(k=5\)=> \(m=\sqrt{\frac{35}{3}}\left(l\right)\)
Vậy \(\left(m,n\right)=\left(5;2\right),\left(-5;7\right)\)