\(10x=15y=6z\Rightarrow x=\frac{3}{2}y;z=\frac{5}{2}y;z=\frac{5}{3}x\Rightarrow10x-5y+z=10y+z=\frac{5}{2}y+10y=\frac{25}{2}y=25\Rightarrow y=2\Rightarrow x=3;z=5\)
a) Ta có : \(10x=15y=6z\)
\(\Rightarrow\frac{10x}{60}=\frac{15y}{60}=\frac{6z}{60}\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{10}\Rightarrow\frac{10x}{60}=\frac{5y}{20}=\frac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\frac{10x}{60}=\frac{5y}{20}=\frac{z}{10}=\frac{10x-5y+z}{60-20+10}=\frac{25}{50}=\frac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\\z=5\end{matrix}\right.\)
Vậy : \(\left(x,y,z\right)=\left(3,2,5\right)\)
c)Đặt \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
\(\Rightarrow xy=2k.3k=54\)
\(\Leftrightarrow6.k^2=54\)
\(\Leftrightarrow k^2=9\)
\(\Leftrightarrow k=\pm9\)
+) Với \(k=9\Rightarrow\left\{{}\begin{matrix}x=2.9=18\\y=3.9=27\end{matrix}\right.\)
+) Với \(k=-9\Rightarrow\left\{{}\begin{matrix}x=2.\left(-9\right)=-18\\y=3.\left(-9\right)=-27\end{matrix}\right.\)
Vậy : \(\left(x,y\right)\in\left\{\left(18,27\right);\left(-18,-27\right)\right\}\)
a) Ta có : 10x=15y=6z10x=15y=6z
⇒10x60=15y60=6z60⇒x6=y4=z10⇒10x60=5y20=z10⇒10x60=15y60=6z60⇒x6=y4=z10⇒10x60=5y20=z10
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
10x60=5y20=z10=10x−5y+z60−20+10=2550=1210x60=5y20=z10=10x−5y+z60−20+10=2550=12
⇒⎧⎪⎨⎪⎩x=3y=2z=5⇒{x=3y=2z=5
Vậy : (x,y,z)=(3,2,5)(x,y,z)=(3,2,5)
c)Đặt x2=y3=kx2=y3=k
⇒{x=2ky=3k⇒{x=2ky=3k
⇒xy=2k.3k=54⇒xy=2k.3k=54
⇔6.k2=54⇔6.k2=54
⇔k2=9⇔k2=9
⇔k=±9⇔k=±9
+) Với k=9⇒{x=2.9=18y=3.9=27k=9⇒{x=2.9=18y=3.9=27
+) Với k=−9⇒{x=2.(−9)=−18y=3.(−9)=−27k=−9⇒{x=2.(−9)=−18y=3.(−9)=−27
Vậy : (x,y)∈{(18,27);(−18,−27)}
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