Toán Lớp 6: Tim x , y ,z ∈ Z , biết :
a) 12/ -6 = x/5 = -y /3 = 2/-17
b) -24/-6 = x/3 = 4/y ² = z ³/-2
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Toán Lớp 6: Tim x , y ,z ∈ Z , biết :
a) 12/ -6 = x/5 = -y /3 = 2/-17
b) -24/-6 = x/3 = 4/y ² = z ³/-2
Comments ( 2 )
Giải đáp:+Lời giải và giải thích chi tiết:
a) 12/ -6 = x/5 = -y /3 = 2/-17
=>[(12/-6=x/5),(-y/3=12/-6):}
=>[(-6x=12.5), (-6.(-y)=12.3):}
=>[(-6x=60), (6y=36):}
=>[(xx=60:(-6)), (y=36:6):}
=>[(x=-10), (y=6):}
Vậy x=-10;y=6
b)-24/-6 = x/3 = 4/y^2 = z^3/-2
Hay 4=x/3=4/y^2=z^3/2
=>[(x.1=4.3), (4y^2=4), (z^3=2.4):}
=>[(x=12),(y^2=1), (z^3=8):}
=>[(x=12), (y=1; y=-1), (z=2):}
Vậy x=12; y in {-1;1}; z=2
$ a) \dfrac{12}{-6} = \dfrac{x}{5} = \dfrac{-y}{3} $
$ ⇒ -2 = \dfrac{x}{5} = \dfrac{-y}{3} $
$ ⇒ $ \(\left[ \begin{array}{l}x=-2.5=-10\\-y=-2.3=-6⇒y=6 \end{array} \right.\)
$ b) \dfrac{-24}{6} = \dfrac{x}{3} = \dfrac{4}{y^2} = \dfrac{z^3}{-2} $
$ ⇒ 4 = \dfrac{x}{3} = \dfrac{4}{y^2} = \dfrac{z^3}{-2} $
$ ⇒ $ \(\left[ \begin{array}{l}x=3.4=12\\y^2=1⇒y=±1\\z^3=-8⇒z=-2\end{array} \right.\)