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Đặt \(\frac{x}{3}=\frac{y}{5}=k\)
\(\Rightarrow x=3k;y=5k\)
Ta có:
\(B=\frac{5\cdot\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{120k^2}{15k^2}=8\)
Vậy B=8
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Đặt \(\frac{x}{3}=\frac{y}{5}=k\left(k≠0\right)\Rightarrow\hept{\begin{cases}x=3k\\y=5k\end{cases}}\Rightarrow A=\frac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\)
\(\Rightarrow A=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{120k^2}{15k^2}=8\left(\text{do k ≠ 0}\right)\)
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\(\frac{x}{3}=\frac{y}{5}\)\(\Rightarrow x=\frac{3y}{5}\)
Thay vào biểu thức A ta được:
\(A=\frac{5.\left(\frac{3y}{5}\right)^2+3y^2}{10.\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\frac{9y^2+15y^2}{5}}{\frac{18y^2-15y^2}{5}}=\frac{24y^2}{3y^2}=8\)
Đặt \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow x=3k,y=5k\)
Ta có: \(A=\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{k^2\left(45+75\right)}{k^2\left(90-75\right)}=\frac{120k^2}{15k^2}=8\)
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Đặt x/3=y/5=k
=>x=3k; y=5k
\(A=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{5\cdot9+3\cdot25}{10\cdot9-3\cdot25}=8\)
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Đặt \(\frac{x}{3}=\frac{y}{5}=k\)=> \(x=3k\) ; \(y=5k\)
Khi đó, ta có: C = \(\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)
= \(\frac{5.3^2.k^2+3.5^2.k^2}{10.3^2.k^2-3.5^2.k^2}\)
= \(\frac{k^2.\left(5.9+3.25\right)}{k^2.\left(10.9-3.25\right)}\)
= 8
\(\frac{x}{3}=\frac{y}{5}\Rightarrow5x=3y\)
\(C=\frac{3xy+5xy}{6xy-5xy}=\frac{8xy}{1xy}=8\)
cách này nhanh hơn không :v
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Ta có: \(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\)
\(\Rightarrow x=3k\)
\(y=5k\)
Khi đó \(P=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)
\(=\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}\)
\(=\dfrac{120k^2}{15k^2}=\dfrac{120}{15}=8.\)
\(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow x=3k;y=5k\)
\(P=\dfrac{5x^2+3y^2}{10x^2-3y^2}\)
\(P=\dfrac{5.3k^2+3.5k^2}{10.3k^2-3.5k^2}\)
\(P=\dfrac{15k^2+15k^2}{30k^2-15k^2}\)
\(P=\dfrac{30k^2}{15k^2}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta gọi: \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow x=3k;y=5k\)
Thay x = 3k và y = 5k vào A ta có: \(A=\frac{5^2.\left(3k\right)^2+3^2.\left(5k\right)^2}{10^2.\left(3k\right)^2-3^2.\left(5k\right)^2}=\frac{25.9k^2+9.25k^2}{100.9k^2-9.25k^2}=\frac{9.25k^2\left(1+1\right)}{9.25k^2\left(4-1\right)}=\frac{2}{3}\)
\(\dfrac{x}{3}=\dfrac{y}{5}\) \(\Leftrightarrow x=\dfrac{3}{5}y\)
Thay vào B có
\(B=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5\cdot\left(\dfrac{3}{5}y\right)^2+3y^2}{10\cdot\left(\dfrac{3}{5}y\right)^2-3y^2}=\dfrac{5\cdot\dfrac{9}{25}y^2+3y^2}{10\cdot\dfrac{9}{25}y^2-3y^2}\)
\(=\dfrac{\dfrac{9}{5}y^2+3y^2}{\dfrac{18}{5}y^2-3y^2}=\dfrac{y^2\left(\dfrac{9}{5}+3\right)}{y^2\left(\dfrac{18}{5}-3\right)}=\dfrac{\dfrac{9}{5}+3}{\dfrac{18}{5}-3}=\dfrac{\dfrac{24}{5}}{\dfrac{3}{5}}\)
\(=\dfrac{24}{5}\cdot\dfrac{5}{3}=8\)