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Lời giải:
a.
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)
b)
Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)
TA có : 4x = 5y
=> \(\frac{x}{5}=\frac{y}{4}=t\)
=> x = 5t ; y = 4t
x^2 - y^2 = 25t^2 - 16t^2 = 1
=> 9t^2 = 1
=> t^2 = 1/9 => t = 1/3 ( vì x ; y dương => t dương )
(+) với t = 1/3 => x = 5.1/3 = 5/3
=> y = 4.1/3 = 4/3
Tích là : 5/3 . 4/3 = 20/9
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
Lời giải:
$4x=5y\Rightarrow x=\frac{5}{4}y$. Khi đó:
$x^2-y^2=1$
$\Rightarrow (\frac{5}{4}y)^2-y^2=1$
$\Rightarrow \frac{25}{16}y^2-y^2=1$
$\Rightarrow \frac{9}{16}y^2=1\Rightarrow y^2=\frac{16}{9}$
$\Rightarrow y=\pm \frac{4}{3}$
Nếu $y=\frac{4}{3}$ thì $x=\frac{5}{4}.\frac{4}{3}=\frac{5}{3}$
$\Rightarrow xy=\frac{4}{3}.\frac{5}{3}=\frac{20}{9}$
Nếu $y=\frac{-4}{3}$ thì $x=\frac{5}{4}.\frac{-4}{3}=\frac{-5}{3}$
$\Rightarrow xy=\frac{-4}{3}.\frac{-5}{3}=\frac{20}{9}$
Vậy $xy=\frac{20}{9}$
4x =5y => \(\frac{x}{5}=\frac{y}{4}=k\)
k2 = \(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)
=> \(k=\frac{-1}{3}\)
hoặc \(k=\frac{1}{3}\)
vì x,y dương nên \(k=\frac{1}{3}\)
vậy x=1/3.5 = 5/3
y=1/3.4 = 4/3
=>xy = 5/3.4/3=20/9