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Bài này dễ thôi
\(x:y:z=3:4:5=>\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
hay \(\frac{3x^2}{27}=\frac{2y^2}{32}=\frac{5z^2}{125}\) và \(5z^2-3x^2-2y^2=594\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{5z^2}{125}=\frac{3x^2}{27}=\frac{2y^2}{32}\) = \(\frac{5z^2-3x^2-2y^2}{125-27-32}\) = \(\frac{594}{66}\) = 9
=> x = 3.9 = 27
y = 4.9 = 36
z = 5.9 = 45
a)Ta có:
\(x:y:z=3:4:5\Leftrightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)
\(\Rightarrow5z^2-3x^2-2y^2=594\Leftrightarrow5\left(5k\right)^2-3\left(3k\right)^2-2\left(4k\right)^2=594\)
\(\Leftrightarrow125k^2-27k^2-32k^2=66k^2=594\Leftrightarrow k^2=9\Leftrightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)TH1:k=3
\(\Rightarrow\left\{{}\begin{matrix}x=3k=9\\y=4k=12\\z=5k=15\end{matrix}\right.\)
TH2:k=-3
\(\Rightarrow\left\{{}\begin{matrix}x=3k=-9\\y=4k=-12\\z=5k=-15\end{matrix}\right.\)
b)Ta có:
\(x+y=3\left(x-y\right)\Leftrightarrow x+y=3x-3y\Leftrightarrow y+3y=3x-x\Leftrightarrow4y=2x\Leftrightarrow2y=x\)
Lại có:
\(x+y=x:y\Leftrightarrow2y+y=2y:y\Leftrightarrow3y=2\Leftrightarrow y=\frac{2}{3}\)
\(\Rightarrow x=2y=2.\frac{2}{3}=\frac{4}{3}\)
a) Ta có :
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Leftrightarrow y=\frac{4x}{3};z=\frac{5x}{3}\)
\(\Rightarrow5\left(\frac{5x}{3}\right)^2-3x^2-2\left(\frac{4x}{3}\right)^2=594\)
\(\Leftrightarrow\frac{125x^2}{9}-\frac{27x^2}{9}-\frac{32x^2}{9}=594\)
\(\Leftrightarrow\frac{66x^2}{9}=594\Leftrightarrow x^2=\frac{594.9}{66}\)
\(\Leftrightarrow x^2=81\Leftrightarrow x=\pm9\)
\(\Leftrightarrow y=\pm12;z=\pm15\)
Vậy . . . . . . .
b) \(x+y=\frac{x}{y}=3\left(x-y\right)\)
\(\Rightarrow x+y+3\left(x-y\right)=\frac{2x}{y}\Leftrightarrow4x-2y=\frac{2x}{y}\left(1\right)\)
\(x+y-3\left(x-y\right)=0\Leftrightarrow4y-2x=0\Leftrightarrow x=2y\)
Thay x = 2y vào pt (1) , ta có :
\(8y-2y=\frac{4y}{y}\Leftrightarrow6y=4\Leftrightarrow y=\frac{2}{3}\)
\(\Rightarrow x=\frac{2.2}{3}=\frac{4}{3}\)
Vậy . . . . . . .
Từ \(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{3x^2}{27}=\dfrac{2y^2}{32}=\dfrac{5z^2}{125}\)
Theo t/c dãy số bằng nhau :
\(\dfrac{3x^2}{27}=\dfrac{2y^2}{32}=\dfrac{5z^2}{125}=\dfrac{5z^2-2y^2-3x^2}{125-32-27}=\dfrac{594}{66}=9\)
\(\Rightarrow3x^2=9\cdot27=243\Rightarrow x^2=\dfrac{243}{3}=81\Rightarrow x\in\left\{9;-9\right\}\)
\(\Rightarrow2y^2=9\cdot32=288\Rightarrow y^2=\dfrac{288}{2}=144\Rightarrow y\in\left\{12;-12\right\}\)
\(\Rightarrow5z^2=9\cdot125=1125\Rightarrow z^2=\dfrac{1125}{5}=225\Rightarrow z\in\left\{15;-15\right\}\)
a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2
vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6
\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8
\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10
vậy x=6,y=8,z=10
vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)
từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1
vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9
\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12
\(\dfrac{z}{16}\)=-1=>z=-1.16=-16
vậy...
\(x:y:z=3:4:5\)
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\) và \(5z^2-3x^2-2y^2\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{5z^2-3x^2-2y^2}{5.5^2-3.3^2-2.4^2}=\frac{594}{66}=9\)
\(\Leftrightarrow\frac{x}{3}=9\Rightarrow x=9.3=27\)
\(\Leftrightarrow\frac{y}{4}=9\Rightarrow y=9.4=36\)
\(\Leftrightarrow\frac{z}{5}=9\Rightarrow z=9.5=45\)
Vậy x = 27 ; y = 36 ; z = 45
\(x+y=3\left(x-y\right)\)
\(\Rightarrow x+y=3x-3y\)
\(\Rightarrow y+3y=3x-x\)
\(\Rightarrow4y=2x\)
\(\Rightarrow2y=x\)
\(\Rightarrow x:y=2\)
\(\Rightarrow x+y=2y+y=2\)
\(\Rightarrow3y=2\)
\(\Rightarrow y=\frac{2}{3}\)
\(\Rightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3};y=\frac{2}{3}\)
a) \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{5z^2}{125}=\frac{3x^2}{27}=\frac{2y^2}{32}=\frac{5z^2-3x^2-2y^2}{125-27-32}=\frac{594}{66}=9\)
\(\Rightarrow5z^2=9.125=1125\Rightarrow z^2=225\Rightarrow z=\pm15\)
\(3x^2=9.27=243\Rightarrow x^2=81\Rightarrow x=\pm9\)
\(2y^2=9.32=288\Rightarrow y^2=144\Rightarrow y=\pm12\)
Vậy ....
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