Đặt \(\frac{x}{3}=\frac{y}{5}=n\Rightarrow x=3n;y=5n\)

\(\Rightarrow A=\frac{5.3^2n^2+3.5^2n^2}{10.3^2n^2-3.5^2n^2}=\frac{n^2\left(45+75\right)}{n^2\left(90-75\right)}=\frac{n^2.120}{n^2.25}=\frac{24}{5}\)

\(\frac{x}{3}=\frac{y}{5}\Rightarrow5x=3y\)

Thay 3y = 5x ; ta được: 

\(A=\frac{5x^2+5x^2}{10x^2-5x^2}=\frac{2\times5x^2}{2\times5x^2-5x^2}=\frac{2\times5x^2}{5x^2\times\left(2-1\right)}=\frac{2\times5x^2}{5x^2\times1}=2\)  

Đặt \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow x=3k;y=5k\)

\(A=\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}=\frac{5.3^2.k^2+3.5^2.k^2}{10.3^2.k^2-3.5^2.k^2}\)

\(A=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{\left(45+75\right).k^2}{\left(90-75\right).k^2}=\frac{120k^2}{15k^2}=\frac{120}{15}=8\)

Vậy A=8
 

1.Tính giá trị của biểu thức: A=\(\frac{5x^2+3y^2}{10x^2-3y^2}\left(1\right)biết\frac{x}{3}=\frac{y}{5}suyra:5x=3y;suyra:x=\frac{3y}{5};thayvào\left(1\right)taco:\frac{5\left(\frac{3y}{5}\right)^2+3y^2}{10\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\frac{9y^2}{5}+3y^2}{\frac{18y^2}{5}-3y^2}=\frac{24y^2}{5}\cdot\frac{5}{3y^2}=8\)

2.\(\frac{x}{y}=\frac{7}{10}suyra;\frac{x}{7}=\frac{y}{10}\left(1\right)và\frac{y}{z}=\frac{5}{8}suyra;\frac{y}{5}=\frac{z}{8}suyra;\frac{y}{5}\cdot\frac{1}{2}=\frac{z}{8}\cdot\frac{1}{2}suyra;\frac{y}{10}=\frac{z}{16}\left(2\right)Tù\left(1\right)và\left(2\right)suyra\frac{x}{7}=\frac{y}{10}=\frac{z}{16}và2x+5y-2z=9;suyra:\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}ápdụngtínhchấtcủadãytỉsốbằngnhautacó\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}=\frac{2x+5y-2z}{14+50-32}=\frac{9}{32}suyra;x=\frac{63}{32};y=\frac{45}{16};z=\frac{9}{2}\)

đặt \(\frac{x}{3}=\frac{y}{5}=k\)(k khac 0)

 Từ  \(\frac{x}{3}=k\)=>x=3k

        \(\frac{y}{5}=k\)=>y=5k

roi ban thay vao tinh

\(a)\)  Ta có : 

\(\frac{x}{18}=\frac{y}{9}\)\(\Leftrightarrow\)\(\frac{x}{2}=y\)

\(\Rightarrow\)\(x=2y\)

Thay \(x=2y\) vào \(A=\frac{2x-3y}{2x+3y}\) ta được : 

\(A=\frac{2.2y-3y}{2.2y+3y}=\frac{4y-3y}{4y+3y}=\frac{y}{7y}=\frac{1}{7}\)

Vậy ... ( tự kết luận ) 

Chúc bạn học tốt ~ 

ỳgyjwegfeukwfhưe

\(\frac{x}{3}=\frac{y}{5}\)\(\Rightarrow x=\frac{3y}{5}\)

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: \(A=\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{k^2\left(45+75\right)}{k^2\left(90-75\right)}=\frac{120k^2}{15k^2}=8\)

\(\frac{x}{3}=\frac{y}{5}\Rightarrow x=\frac{3y}{5}\)

\(\Rightarrow B=\frac{5\left(\frac{3y}{5}\right)^2+3y^2}{10\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\left(\frac{9}{5}+3\right)y^2}{\left(\frac{18}{5}-3\right)y^2}=\frac{\frac{9}{5}+3}{\frac{18}{5}-3}=8\)

bn ở đội tuyển toán à

Giải:
Đặt \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow\left\{\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)

Ta có: \(B=\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}=\frac{45.k^2+75k^2}{90k^2-75k^2}=\frac{\left(45+75\right)k^2}{\left(90-75\right)k^2}\)

\(=\frac{120k^2}{15k^2}=\frac{120}{15}=8\)

Vậy B = 8