Tìm x, y, z biết
\(\dfrac{x-1}{2}\)=\(\text{}\dfrac{y-2}{3}\)=\(\dfrac{z-3}{4}\)và \(x+y+z=-34\)
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\(3^{12}=3^{3.4}=\left(3^3\right)^4=27^4\)
\(5^8=5^{2.4}=\left(5^2\right)^4=25^4\)
Do \(27^4>25^4\) nên \(3^{12}>5^8\)
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Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{x+y+z-1-2-3}{2+3+4}=\dfrac{-34-6}{9}=\dfrac{-40}{9}\)
=>\(\left\{{}\begin{matrix}x-1=-\dfrac{40}{9}\cdot2=-\dfrac{80}{9}\\y-2=-\dfrac{40}{9}\cdot3=-\dfrac{120}{9}\\z-3=-\dfrac{40}{9}\cdot4=-\dfrac{160}{9}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{80}{9}+1=-\dfrac{71}{9}\\y=-\dfrac{120}{9}+2=-\dfrac{120}{9}+\dfrac{18}{9}=-\dfrac{102}{9}\\z=-\dfrac{160}{9}+3=-\dfrac{160}{9}+\dfrac{27}{9}=-\dfrac{133}{9}\end{matrix}\right.\)