Tìm x,y,z biết
x/y=-7/4 và 4x—5y=72
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Ta có: \(\frac{x}{y}=\frac{-7}{4}\Rightarrow\frac{x}{-7}=\frac{y}{4}\)
Suy ra \(\frac{4x}{-28}=\frac{5y}{20}\)
Áp dụng tính chất dãy các tỉ số bằng nhau, ta có:
\(\frac{4x}{-28}=\frac{5y}{20}=\frac{4x-5y}{-28-20}=\frac{-3}{2}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-3}{2}.\left(-7\right)=\frac{21}{2}\\y=\frac{-3}{2}.4=-6\end{cases}}\)
Vậy \(x=\frac{21}{2}\) và y = -6
đặt \(\frac{x}{-3}=\frac{y}{8}=k\) \(\Rightarrow x=-3k;y=8k\)
\(x^2-y^2=-\frac{44}{5}\)\(\Leftrightarrow\left(-3k\right)^2-\left(8k\right)^2=9k^2-64k^2=-55k^2=\frac{-44}{5}\)
\(\Rightarrow k^2=\frac{4}{25}\Rightarrow k=\pm\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-6}{5};y=\frac{16}{5}\\x=\frac{6}{5};y=\frac{-16}{5}\end{cases}}\)
\(\Leftrightarrow xy=63\)
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)
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:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
\(\frac{x}{y}=\frac{7}{4}\Rightarrow\frac{x}{7}=\frac{y}{4}=\frac{4x}{28}=\frac{5y}{20}\) Áp dụng TC DTSBN ta có :
\(\frac{4x}{28}=\frac{5y}{20}=\frac{4x-5y}{28-20}=\frac{72}{8}=9\)
=> x = 63 ; y = 36
\(\frac{x}{y}=\frac{-7}{4}\Rightarrow4x=-7y\Rightarrow4x+7y=0\)
Ta có: \(4x-5y=72\Rightarrow4x+7y-12y=72\)
Thay 4x+7y=0 vào đẳng thức,ta có:\(0-12y=72\Rightarrow12y=-72\Rightarrow y=-6\)
\(\Rightarrow x=\frac{-7\times-6}{4}=10.5\)
có Z đâu pn
nếu ko có Z thì X = 10,5; Y = -6