Tìm x, y, z
a, \(x=\dfrac{y}{2}=\dfrac{z}{3}\) và 4x-3y+2z=36
b, \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}\) và 2x-3z=44
c, \(\dfrac{-x}{7}=\dfrac{y}{11}=\dfrac{-z}{5}\) và -3z- 2y-x= -88
d,\(\dfrac{y}{12}=\dfrac{-x}{5}=\dfrac{z}{11}\) và 5y- 2z=114
e, \(\dfrac{z}{32}=\dfrac{y}{17}=\dfrac{x}{25}\)và -2x+ 3y- 4x= -452
a)Xét \(x=\dfrac{y}{2}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=k\\y=2k\\z=3k\end{matrix}\right.\) (1)
Thay (1) vào 4x - 3y + 2z = 36
\(\Rightarrow4.k-3.2k+2.3k=36\)
\(\Rightarrow4k-6k+6k=36\Rightarrow4k=36\)
\(\Rightarrow k=\dfrac{36}{4}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=2.4=8\\z=3.4=12\end{matrix}\right.\)
Vậy...............................................................
b) Xét \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=7k\end{matrix}\right.\) (2)
Thay (2) vào 2x - 3z = 44
\(\Rightarrow2.5k-3.7k=44\)
\(\Rightarrow-11k=44\Rightarrow k=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.\left(-4\right)=-20\\y=4.\left(-4\right)=-16\\z=7.\left(-4\right)=-28\end{matrix}\right.\)
Vậy,................................................
c) Xét \(\dfrac{-x}{7}=\dfrac{y}{11}=\dfrac{-z}{5}=\dfrac{x}{-7}=\dfrac{z}{-5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-7k\\y=11k\\z=-5k\end{matrix}\right.\) (3)
Thay (3) vào -3z - 2y - x = -88
\(\Rightarrow-3.\left(-5k\right)-2.11k-\left(-7k\right)=-88\)
\(\Rightarrow15k-22k+7k=-88\Rightarrow0k=88\)
\(\Rightarrow k\in\varnothing\)
Suy ra: Không có cặp ( x; y; z) thỏa mãn
Vậy.................................................................
d) Xét \(\dfrac{y}{12}=\dfrac{x}{-5}=\dfrac{z}{11}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5k\\y=12k\\z=11k\end{matrix}\right.\) (4)
Thay (4) vào 5y - 2z = 114
\(\Rightarrow6.12k-2.11k=114\)
\(\Rightarrow50k=114\Rightarrow k=2,28\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5.2,28=-11,4\\y=12.2,28=27,36\\z=25,08\end{matrix}\right.\)
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e) Xét \(\dfrac{x}{25}=\dfrac{y}{17}=\dfrac{z}{32}=k\)
\(\left\{{}\begin{matrix}x=25k\\y=17k\\z=32k\end{matrix}\right.\) (5)
Thay (5) vào -2z + 3y - 4x = -452
\(\Rightarrow\left(-2\right).32k+3.17k-4.25k=-452\)
\(\Rightarrow-113k=-452\Rightarrow k=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=25.5=100\\y=17.4=68\\z=32.4=128\end{matrix}\right.\)
Vậy.......................................................
a) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(x=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{4x}{4}-\dfrac{3y}{6}+\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{36}{4}=9\)
+) \(\dfrac{x}{1}=9\Rightarrow x=9\)
+) \(\dfrac{y}{2}=9\Rightarrow y=18\)
+) \(\dfrac{z}{3}=9\Rightarrow z=27\)
Vậy x = 9; y = 18; z = 27.
tương tự
a)
+) Ta có: \(x=\dfrac{y}{2}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=k\\y=2k\\z=3k\end{matrix}\right.\)
+) Đặt: \(4x-3y+2z=36\)
\(\Rightarrow4.k+3.2k-2.3k=36\)
\(\Rightarrow4k+6k-6k=36\)
\(\Rightarrow4k=36\\\)
\(\Rightarrow k=36:4\)
\(\Rightarrow k=9\)
+) Thay vào cách đặt: \(\left\{{}\begin{matrix}x=9\\y=2.9\\z=3.9\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}x=9\\y=18\\z=27\end{matrix}\right.\)
b)
+) Ta có: \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=7k\end{matrix}\right.\)
+) Đặt: \(2x-3z=44\)
\(\Rightarrow2.5k-3.7k=44\)
\(\Rightarrow-11k=44\)
\(\Rightarrow k=44:\left(-11\right)\)
\(\Rightarrow k=-4\)
+) Thay vào cách đặt :\(\left\{{}\begin{matrix}x=5.\left(-4\right)\\y=4.\left(-4\right)\\z=7.\left(-4\right)\end{matrix}\right.\)
+) Vậy : \(\left\{{}\begin{matrix}x=-20\\y=-16\\z=-28\end{matrix}\right.\)
d)
+) Ta có: \(\dfrac{y}{12}=\dfrac{-x}{5}=\dfrac{z}{11}=k\Rightarrow\left\{{}\begin{matrix}y=12k\\x=-5k\\z=11k\end{matrix}\right.\)
* \(\dfrac{-x}{5}\) đổi ngược lại thành \(\dfrac{x}{-5}\)
+) Đặt: \(5y-2z=114\)
\(\Rightarrow5.12k-2.11k=114\)
\(\Rightarrow38k=114\)
\(\Rightarrow k=114:38\)
\(\Rightarrow k=3\)
+) Thay vào cách đặt: \(\left\{{}\begin{matrix}y=12.3\\x=\left(-5\right).3\\z=11.3\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}y=36\\x=-15\\z=33\end{matrix}\right.\)
Mình giải vậy thôi bạn nhá!!!