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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...
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Ta có:
\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)
Chúc bạn học tốt!
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b, \(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
áp dụng dãy tỉ số bằng nhau :
\(\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
x = 2 . 10 = 20
y = 2 . 15 = 30
z = 2 . 21 = 42
Vậy : .....
a, \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)
MSC của y là : 20
Có: \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Áp dụng dãy tỉ số bằng nhau, ta có:
\(2x+3y-z=186\)
\(\Rightarrow2.15+3.20-28=30+60-28=62\)
\(\frac{186}{62}=3\)
x = 3 . 15 = 45
y = 3 . 20 = 60
z = 3 . 28 = 84
Vậy: .....
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x : y : z : t = 2 : 3 : 4 : 5
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{t}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{t}{5}=\frac{x+y+z+t}{2+3+4+5}=\frac{2}{7}\)
\(\Rightarrow x=\frac{2}{7}.2=\frac{4}{7};y=\frac{2}{7}.3=\frac{6}{7};z=\frac{2}{7}.4=\frac{8}{7};t=\frac{2}{7}.5=\frac{10}{7}\)
Ta có: \(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{12}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{49}{7}=7\)
\(\Rightarrow x=7.10=70;y=7.15=105;z=7.12=84\)
x- y/3=x+ y/13 =>x-x=y/3+y/13 => 0=y/3+ y/13 => y/-3=y/13
=> y=0 => x bằng tất cả mọi số