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\(\dfrac{x}{3}=\dfrac{y-5}{7}=\dfrac{z+2}{3}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}\)
\(x+2y=5z\Leftrightarrow x+2y-5z=0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}=\dfrac{x+2y-10-5z-10}{3+14-15}\)
\(=\dfrac{-20}{2}=-10\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-30\\y=-65\\z=-32\end{matrix}\right.\)
Vậy...
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
\(\Rightarrow6.12k-2.11k=114\)
\(\Rightarrow50k=114\Rightarrow k=2,28\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5.2,28=-11,4\\y=12.2,28=27,36\\z=25,08\end{matrix}\right.\)
Vậy..............................................
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,
\(\dfrac{2x}{3y}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{2x}{-1}=\dfrac{3y}{3}\\ \Leftrightarrow\dfrac{-2x}{1}=\dfrac{3y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-2x}{1}=\dfrac{3y}{3}=\dfrac{-2x+3y}{1+3}=\dfrac{7}{4}\)
\(\dfrac{-2x}{1}=\dfrac{7}{4}\Rightarrow-2x=\dfrac{7}{4}\Rightarrow x=\dfrac{7}{4}:\left(-2\right)=\dfrac{-7}{8}\\ \dfrac{3y}{3}=\dfrac{7}{4}\Rightarrow y=\dfrac{7}{4}\)
Vậy \(x=\dfrac{-7}{8};y=\dfrac{7}{4}\)
b,
\(\dfrac{x}{3}=\dfrac{y}{4}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{5y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{5y}{20}=\dfrac{2x+5y}{6+20}=\dfrac{10}{26}=\dfrac{5}{13}\\ \dfrac{x}{3}=\dfrac{2x}{6}=\dfrac{5}{13}\Rightarrow x=\dfrac{5}{13}\cdot3=\dfrac{15}{13}\\ \dfrac{y}{4}=\dfrac{5y}{20}=\dfrac{5}{13}\Rightarrow y=\dfrac{5}{13}\cdot4=\dfrac{20}{13}\)
Vậy \(x=\dfrac{15}{13};y=\dfrac{20}{13}\)
c,
\(7x=3y\\ \Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \dfrac{x}{3}=-4\Rightarrow x=\left(-4\right)\cdot3=-12\\ \dfrac{y}{7}=-4\Rightarrow y=\left(-4\right)\cdot7=-28\)
Vậy \(x=-12;y=-28\)
d,
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}=\dfrac{x+y+\left(-2z\right)}{5+1+4}=\dfrac{x+y-2z}{10}=\dfrac{160}{10}=16\\ \dfrac{x}{5}=16\Rightarrow x=16\cdot5=80\\ \dfrac{y}{1}=16\Rightarrow y=16\\ \dfrac{z}{-2}=\dfrac{-2z}{4}=16\Rightarrow z=16\cdot\left(-2\right)=-32\)
Vậy \(x=80;y=16;z=-32\)
e,
\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\\ \Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\\ \Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)
\(\dfrac{x}{20}=\dfrac{2x}{40}=\dfrac{33}{7}\Rightarrow x=\dfrac{33}{7}\cdot20=\dfrac{660}{7}\\ \dfrac{y}{10}=\dfrac{3y}{30}=\dfrac{33}{7}\Rightarrow y=\dfrac{33}{7}\cdot10=\dfrac{330}{7}\\ \dfrac{z}{15}=\dfrac{4z}{60}=\dfrac{33}{7}\Rightarrow z=\dfrac{33}{7}\cdot15=\dfrac{495}{7}\)
Vậy \(x=\dfrac{660}{7};y=\dfrac{330}{7};z=\dfrac{495}{7}\)
f,
\(\dfrac{x}{-2}=\dfrac{-y}{4}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}=\dfrac{x+\left(-2y\right)+3z}{\left(-2\right)+8+15}=\dfrac{x-2y+3z}{21}=\dfrac{1200}{21}=\dfrac{400}{7}\)
\(\dfrac{x}{-2}=\dfrac{400}{7}\Rightarrow x=\dfrac{400}{7}\cdot\left(-2\right)=\dfrac{-800}{7}\\ \dfrac{-y}{4}=\dfrac{-2y}{8}=\dfrac{400}{7}\Rightarrow-y=\dfrac{400}{7}\cdot4=\dfrac{1600}{7}\Rightarrow y=\dfrac{-1600}{7}\\ \dfrac{z}{5}=\dfrac{3z}{15}=\dfrac{400}{7}\Rightarrow z=\dfrac{400}{7}\cdot5=\dfrac{2000}{7}\)
Vậy \(x=\dfrac{-800}{7};y=\dfrac{-1600}{7};z=\dfrac{2000}{7}\)
g,
\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}=\dfrac{2x+3y-z}{6+24-5}=\dfrac{50}{25}=2\)
\(\dfrac{x}{3}=\dfrac{2x}{6}=2\Rightarrow x=2\cdot3=6\\ \dfrac{y}{8}=\dfrac{3y}{24}=2\Rightarrow y=2\cdot8=16\\ \dfrac{z}{5}=2\Rightarrow z=2\cdot5=10\)
Vậy \(x=6;y=16;z=10\)
Làm gấp nên k có kiểm tra, bn bấm máy tính dò lại nhé
a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)
Theo đề bài, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)
\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)
Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))
\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)
Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)
\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)
\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)
Vậy x=60; y=72; z=63
TH1: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(y+z=-x\)
\(x+z=-y\)
\(\Rightarrow M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\dfrac{-xyz}{8xyz}=\dfrac{-1}{8}\)
TH2: \(x+y+z\ne0\)
\(\Rightarrow2x+2y-z=3\)
\(\Rightarrow2x+2y=4z\)
\(\Rightarrow x+y=2z\)
\(x+z=2y\)
\(y+z=2x\)
\(\Rightarrow M=\dfrac{2z.2y.2x}{8xyz}=1\)
Vậy: \(M=\dfrac{-1}{8}\) hoặc \(1\)
Ta có \(\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x+2y-z}{z}=3\\\dfrac{2x+2z-y}{y}=3\\\dfrac{2y+2z-x}{x}=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+2y-z=3z\\2x+2z-y=3y\\2y+2z-x=3x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+2y=4z\\2x+2z=4y\\2y+2z=4x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=2z\\x+z=2y\\y+z=2x\end{matrix}\right.\)
Ta có \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}\)
\(\Rightarrow M=\dfrac{2x.2y.2z}{8xyz}=\dfrac{8xyz}{8xyz}=1\)
Vậy \(M=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+2y-z}{z}=\dfrac{2x-y+2z}{y}=\dfrac{-x+2y+2z}{x}=\dfrac{2x+2y-z+2x-y+2z-x+2y+2x}{x+y+z}=\dfrac{3x+3y+3z}{x+y+z}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)
\(\Rightarrow\)\(\dfrac{2x+2y-z}{z}=3\Leftrightarrow2x+2y-z=3z\Leftrightarrow2\left(x+y\right)=4z\Leftrightarrow x+y=2z\Leftrightarrow z=\dfrac{x+y}{2}\)
Tương tự: \(x=\dfrac{y+z}{2}\)
\(y=\dfrac{x+z}{2}\)
Thay vào M, ta được:
\(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\left(\dfrac{y+z}{2}.\dfrac{x+z}{2}.\dfrac{x+y}{2}\right).8}\)
\(=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8}.8}=1\)
Lời giải:
Từ \(2x^2+2y^2-2z^2=-100\)
\(\Leftrightarrow x^2+y^2-z^2=-50\)
\(\Rightarrow z^2=x^2+y^2+50\)
Có: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}(*)\Rightarrow \left(\frac{x}{3}\right)^2=\left(\frac{y}{4}\right)^2=\left(\frac{z}{5}\right)^2\)
\(\Leftrightarrow \frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\Leftrightarrow \frac{x^2+y^2}{9+16}=\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{x^2+y^2+50}{25}\)
\(\Leftrightarrow \frac{x^2+y^2}{25}=\frac{x^2+y^2+50}{25}\Leftrightarrow x^2+y^2=x^2+y^2+50\)
\(\Leftrightarrow 0=50\) (vô lý)
Do đó pt vô nghiệm.
Từ 2x2+2y2−2z2=−1002x2+2y2−2z2=−100
⇔x2+y2−z2=−50⇔x2+y2−z2=−50
⇒z2=x2+y2+50⇒z2=x2+y2+50
Có: x3=y4=z5(∗)⇒(x3)2=(y4)2=(z5)2x3=y4=z5(∗)⇒(x3)2=(y4)2=(z5)2
⇔x29=y216=z225⇔x29=y216=z225
Áp dụng tính chất dãy tỉ số bằng nhau:
⇔x2+y29+16=x29=y216=z225=x2+y2+5025⇔x2+y29+16=x29=y216=z225=x2+y2+5025
⇔x2+y225=x2+y2+5025⇔x2+y2=x2+y2+50⇔x2+y225=x2+y2+5025⇔x2+y2=x2+y2+50
⇔0=50⇔0=50 (vô lý)
Do đó pt vô nghiệm.