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\(\frac{x+y}{13}=\frac{x-y}{3}=\frac{xy}{200}\)
\(\Rightarrow3\left(x+y\right)=13\left(x-y\right)\)
\(\Rightarrow y=\frac{5x}{8}\)
\(\frac{x-y}{3}=\frac{xy}{200}\Rightarrow200\left(x-y\right)=3xy\)
\(\Rightarrow200\left(x-\frac{5x}{8}\right)=\frac{3x.5x}{8}\Rightarrow x^2-40x\Rightarrow x\left(x-40\right)=0\)
\(\Rightarrow x=\left[\begin{array}{nghiempt}0\\40\end{array}\right.\)\(\Leftrightarrow y=\left[\begin{array}{nghiempt}0\\25\end{array}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{13+3}=\frac{2x}{16}=\frac{x}{8}\)
\(\frac{xy}{200}=\frac{x}{8}=\frac{25x}{200}
\)
\(\Rightarrow xy=25x\Rightarrow y=25\)
Ta có : \(\frac{x+y}{13}=\frac{x-y}{3}\)
hay \(\frac{x+25}{13}=\frac{x-25}{3}\)
\(\Rightarrow3x+75=13x-325\)
\(-10x=-400\)
\(x=-400:-10=40\)
Vậy x = 40
y = 25
100% đúng đấy bạn
+x =0 => y =0
+\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x+y-x+y}{13-3}=\frac{2y}{10}=\frac{y}{5}=\frac{xy}{5x}=\frac{xy}{200}\)
=> 5x =200 => x =40
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{2x}{16}=\frac{x}{8}=\frac{y}{5}\)
=> y =5.40/8 = 25
Vậy (x;y) = (0;0); (40;25)
Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}\)
=> (x - y).13 = 3.(x + y)
=> 13x - 13y = 3x + 3y
=> 13x - 3x = 3y + 13y
=> 10x = 16y
=> \(x=\frac{16}{10}y=\frac{8}{5}y\)
Thay \(x=\frac{8}{5}y\) vào đề bài ta có: \(\frac{\frac{8}{5}y-y}{3}=\frac{\frac{8}{5}y+y}{13}=\frac{\frac{8}{5}y.y}{200}\)
\(\Rightarrow\frac{\frac{3}{5}y}{3}=\frac{\frac{13}{5}y}{13}=\frac{\frac{8}{5}y^2}{200}\)
\(\Rightarrow\frac{3}{5}y.\frac{1}{3}=\frac{13}{5}y.\frac{1}{13}=\frac{8}{5}y^2.\frac{1}{200}\)
\(\Rightarrow\frac{1}{5}y=\frac{1}{125}.y^2\)
\(\Rightarrow\frac{1}{5}y-\frac{1}{125}.y^2=0\)
\(\Rightarrow\frac{1}{5}y.\left(1-\frac{1}{25}y\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}y=0\\1-\frac{1}{25}y=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}y=0\\\frac{1}{25}y=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}y=0\\y=25\end{array}\right.\)
+ Với y = 0 thì \(x=\frac{8}{5}.0=0\)
+ Với y = 25 thì \(x=\frac{8}{5}.25=40\)
Vậy \(\begin{cases}x=0\\y=0\end{cases}\); \(\begin{cases}x=40\\y=25\end{cases}\)
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{x}{8}=\frac{25x}{200}=\frac{xy}{200}\)
Vì 25x = xy nên y = 25
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y-x-y}{3-13}=\frac{-2y}{-10}=\frac{y}{5}\)
=> \(\frac{y}{5}=\frac{x}{8}\Rightarrow8y=5x\Rightarrow\frac{x}{y}=\frac{8}{5}\)
=> x = 40
Vậy...
Nguyễn Hải Đăng chắc bn giỏi nói ng ta ngu :((
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y-x-y}{3-13}=\frac{-2y}{-10}=\frac{y}{5}\)
\(\Rightarrow\frac{y}{5}=\frac{xy}{200}\Rightarrow200y=5xy\Rightarrow\frac{200y}{5y}=x\Rightarrow x=40\)
\(\frac{x-y}{3}=\frac{y}{5}=\frac{40-y}{3}=\frac{y}{5}\Rightarrow5.\left(40-y\right)=3y\Rightarrow200-5y=3y\)
\(\Rightarrow200=8y\Rightarrow y=25\)
Vậy x=40, y=25
Ta có :
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{25x}{200}=\frac{xy}{200}\)
\(\Rightarrow25x=xy\Rightarrow y=25\)
\(\Rightarrow\frac{x-25}{3}=\frac{x+25}{13}\)
\(\Leftrightarrow13x-325=3x+75\)
\(\Leftrightarrow10x=400\Rightarrow x=40\)
Vậy \(x=40;y=25\)
Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{xy}{200}\left(1\right)\)
\(\Rightarrow\frac{x-y}{3}=\frac{x+y}{13}=\frac{xy}{200}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{x}{8}\left(2\right)\)
Từ (1) và (2) => \(\frac{x}{8}=\frac{xy}{200}\Rightarrow8xy=200x\)
\(\Leftrightarrow8xy-200x=0\)
\(\Leftrightarrow8x.\left(y-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}8x=0\\y-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\y=25\end{cases}}}\)
* Nếu x = 0 thì \(\frac{0-y}{3}=\frac{0+y}{13}=0\Rightarrow y=0\)
* Nếu y = 25 thì \(\frac{x-25}{3}=\frac{x+25}{13}\)
\(\Leftrightarrow13.\left(x-25\right)=3.\left(x+25\right)\)
\(\Leftrightarrow13x-325=3x+75\)
\(\Rightarrow13x-3x=75+325=400\)
\(\Rightarrow10x=400\)
\(\Rightarrow x=40\)
Vậy x =0 thì y =0
x =40 thì y = 25
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