Ta có :
\(x:y:z:t=2:3:4:5\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}=\dfrac{x+y+z+t}{2+3+4+5}=\dfrac{-42}{14}=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-4\\\dfrac{y}{3}=-4\\\dfrac{z}{4}=-4\\\dfrac{t}{5}=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=-12\\z=-16\\t=-20\end{matrix}\right.\)
Vậy ..
Ta có: x:y:z:t=2:3:4:5\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}\) = \(\dfrac{x+y+z+t}{2+3+4+5}\) = \(\dfrac{42}{14}=3\)
\(\Rightarrow x=2.3=6\)
\(\Rightarrow y=3.3=9\)
\(\Rightarrow z=4.3=12\)
\(\Rightarrow t=5.3=15\)
Vậy x=6 ; y = 9; z =12; t=15
Ta có:\(x:y:z:t=2:3:4:5\) và \(x+y+z+t=-42\)
⇔\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}\) và \(x+y+z+t=-42\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}=\dfrac{x+y+z+t}{2+3+4+5}=\dfrac{-42}{14}=-3\)
Do đó:
\(\dfrac{x}{2}=-3\Leftrightarrow x=-6\)
\(\dfrac{y}{3}=-3\Leftrightarrow y=-9\)
\(\dfrac{z}{4}=-3\Leftrightarrow z=-12\)
\(\dfrac{t}{5}=-3\Leftrightarrow t=-15\)
Vậy x=-6; y=-9; z=-12; t=-15
Ta có: x:y:z:t\(=\)2:3:4:5\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}\)
Theo tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}=\dfrac{x+y+z+t}{2+3+4+5}=\dfrac{-42}{14}=-3\)
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\Rightarrow x=2.\left(-3\right)=-6\\\dfrac{y}{3}=-3\Rightarrow y=3.\left(-3\right)=-9\\\dfrac{z}{4}=-3\Rightarrow z=4.\left(-3\right)=-12\\\dfrac{t}{5}=-3\Rightarrow t=5.\left(-3\right)=-15\end{matrix}\right.\)
Vậy \(x=-6;y=-9;z=-12;t=-15\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{5}=\dfrac{x+y+z+t}{2+3+4+5}=\dfrac{-42}{14}=-3\)
x=-6
y=-9
z=-12
t=-15
