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Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
\(C=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}=\dfrac{120k^2}{15k^2}=8\)
Vậy C = 8
Đặt:
\(\dfrac{x}{3}=\dfrac{y}{5}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
Thay vào \(C\) ta có:
\(C=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}=\dfrac{120k^2}{15k^2}=\dfrac{120}{15}=8\)
\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
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 \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow x=3k;y=5k\)
Thay x=3k;y=5k vào biểu thức C(x;y) ta có:
\(C\left(x;y\right)=\dfrac{5\left(3k\right)^2+3.\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\)
\(=\dfrac{5.9.k^2+3.25.k^2}{10.9.k^2-3.25.k^2}\)
\(=\dfrac{45k^2+75k^2}{90k^2-75k^2}\)
\(=\dfrac{120k^2}{15k^2}=\dfrac{120}{15}=8\)
Vậy giá trị của biểu thức C(x;y) là 8
Chúc bạn học học tốt nha!!!
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
Từ \(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
Khi đó \(P=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5\cdot\left(3k\right)^2+3\cdot\left(5k\right)^2}{10\cdot\left(3k\right)^2-3\cdot\left(5k\right)^2}\)
\(=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}\)
\(=\dfrac{120k^2}{15k^2}=\dfrac{120}{15}=8\)
Ta có:
x/3=y/5
=> x=3/5y
Thay x vào P ta được P