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![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Ta có: \(\frac{x+2}{y+3}=\frac{2}{3}\Rightarrow3\left(x+2\right)=2\left(y+3\right)\)
\(\Rightarrow3x+6=2y+6\)
\(\Rightarrow3x=2y\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow x=2k,y=3k\)
Lại có: \(A=\frac{x^2+y^2}{xy}=\frac{\left(2k\right)^2+\left(3k\right)^2}{2k3k}=\frac{4k^2+9k^2}{6k^2}=\frac{\left(4+9\right)k^2}{6k^2}=\frac{13}{6}\)
Vậy \(A=\frac{13}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x, y tỉ lệ nghịch vs 2, 3
=> 2.x=3.y=> \(x=\frac{3}{2}y\)
y, z tỉ lệ thuận với 4, 3
=> \(\frac{y}{4}=\frac{z}{3}\Rightarrow z=\frac{3}{4}y\)
Em thay vào tính nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\orbr{\begin{cases}2x-3y=3\\x+2y=2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}4x-6y=6\\3x+6y=6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}7x=12\\3x+6y=6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{12}{7}\\3x+6y=6\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{12}{7}\\y=\frac{1}{7}\end{cases}}\)
Vậy tỉ lệ thức \(\frac{y}{x}=\frac{1}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(2x-3y\right).3=\left(x+2y\right).2\)
\(2x.3-3y.3=x.2+2y.2\)
\(6x-9y=2x+4y\)
\(6x-2x=9y+4y\)
\(4x=13y\)
Chia hai vế cho 4y, ta có:
\(\frac{4x}{4y}=\frac{13y}{4y}\)
\(=\frac{x}{y}=\frac{13}{4}\)
\(\Rightarrow\) Tỉ số giữa x và y là \(\frac{13}{4}\)
Ta có:
\(\frac{2x-3y}{x+2y}=\frac{2}{3}\)
<=> 3(2x-3y)= 2(x+2y)
<=> 6x - 9y = 2x + 4y
<=> 6x - 2x = 9y+4y
<=> 4x = 13y
=> \(\frac{x}{y}=\frac{13}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
Áp dụng t/ c của dãy tỉ số bằng nhau ta có: \(\frac{x+2}{y+3}=\frac{2}{3}=\frac{\left(x+2\right)-2}{\left(y+3\right)-3}=\frac{x}{y}\)
=> x = 2k; y = 3k (k khác 0)
=> A = \(\frac{13.\left(3k\right)^2-9.\left(2k\right)^2}{9.\left(3k\right)^2}=\frac{81k^2}{81k^2}=1\)