Đặt x/3=y/5=k

=>x=3k; y=5k

\(A=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{5\cdot9+3\cdot25}{10\cdot9-3\cdot25}=8\)

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:

\(C=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}=\dfrac{120k^2}{15k^2}=\dfrac{120}{15}=8\)

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!!!

Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\) (k \(\ne\) 0)

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

Mà A = \(\dfrac{5x^2+3y^2}{10x^2-3y^2}\) (bài cho)

\(\Rightarrow\) A = \(\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\)

\(\Leftrightarrow\) A = \(\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}\)

\(\Leftrightarrow\) A = \(\dfrac{45k^2+75k^2}{90k^2-75k^2}\)

\(\Leftrightarrow\) A = \(\dfrac{120k^2}{15k^2}\)

\(\Leftrightarrow\) A = \(\dfrac{120}{15}\)

\(\Leftrightarrow\) A = 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

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

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

Ta có:

\(P=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)

\(=\dfrac{45k^2+75k^2}{90k^2-75k^2}=\dfrac{15k^2\left(3+5\right)}{15k^2\left(6-5\right)}=\dfrac{3+5}{6-5}=8\)

Vậy \(P=8\)

Ta có : 

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

Đặt \(\frac{x}{3}=\frac{y}{5}=k\)

\(\Rightarrow\)\(x=3k\)

\(\Rightarrow\)\(y=5k\)

Thay \(x=3k\) và \(y=5k\) vào biểu thức \(\frac{5x^2+3y^2}{10x^2-3y^2}\) ta được : 

\(\frac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\)

\(=\)\(\frac{5.3^2k^2+3.5^2k^2}{10.3^2k^2-3.5^2k^2}\)

\(=\)\(\frac{45k^2+75k^2}{90k^2-75k^2}\)

\(=\)\(\frac{k^2\left(45+75\right)}{k^2\left(90-75\right)}\)

\(=\)\(\frac{45+75}{90-75}\)

\(=\)\(\frac{120}{15}\)

\(=\)\(8\)

Vậy giá trị của biểu thức \(\frac{5x^2+3y^2}{10x^2-3y^2}=8\) khi \(5x=3y\)

Chúc bạn học tốt ~ 

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

thay x vào biểu thức ta được

 \(\frac{5\left(\frac{3y}{5}\right)^2+3y}{10\left(\frac{3y}{5}\right)^2-3y}=\frac{3y^2\left(\frac{1}{5}+1\right)}{3y^2\left(\frac{2}{5}-1\right)}\)

\(=\frac{6}{5}:\left(-\frac{3}{5}\right)=\frac{6}{5}.\left(\frac{5}{-3}\right)\)

\(=-2\)

1) \(4x^5y^2-8x^4y^2+4x^3y^2\)

\(=4x^3y^2\left(x^2-2x+1\right)\)

\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=4x^3y^2\left(x-1\right)^2\)

2) \(5x^4y^2-10x^3y^2+5x^2y^2\)

\(=5x^2y^2\left(x^2-2x+1\right)\)

\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=5x^2y^2\left(x-1\right)^2\)

3) \(12x^2-12xy+3y^2\)

\(=3\left(4x^2-4xy+y^2\right)\)

\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

\(=3\left(2x-y\right)^2\)

4) \(8x^3-8x^2y+2xy^2\)

\(=2x\left(4x^2-4xy+y^2\right)\)

\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

\(=2x\left(2x-y\right)^2\)

5) \(20x^4y^2-20x^3y^3+5x^2y^4\)

\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)

\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

\(=5x^2y^2\left(2x-y\right)^2\)

1: 4x^5y^2-8x^4y^2+4x^3y^2

=4x^3y^2(x^2-2x+1)

=4x^3y^2(x-1)^2

2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)

3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)

4: \(=2x\left(4x^2-4xy+y^2\right)=2x\left(2x-y\right)^2\)

5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)