Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) => \(\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{5}\right)^2\)
=> \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{2x^2+y^2-z^2}{2.9+16-25}=\dfrac{9}{18+16-25}=\dfrac{9}{9}=1\)
=> \(\left\{{}\begin{matrix}\dfrac{x^2}{9}=1\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3\\\dfrac{y^2}{16}=1\Rightarrow\dfrac{y}{4}=1\Rightarrow y=4\\\dfrac{z^2}{25}=1\Rightarrow\dfrac{z}{5}=1\Rightarrow z=5\end{matrix}\right.\)
Vậy x = 3, y = 4, z = 5
\(\dfrac{2x}{5}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Rightarrow\dfrac{2}{5}x=\dfrac{3}{4}y=\dfrac{4}{5}z\)
\(\Rightarrow\dfrac{2}{5}x.\dfrac{1}{12}=\dfrac{3}{4}y.\dfrac{1}{12}=\dfrac{4}{5}z.\dfrac{1}{12}\)
\(\Rightarrow\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=30k\\y=16k\\z=15k\end{matrix}\right.\). Ta có:
\(x+y+z=49\)
\(\Rightarrow30k+16k+15k=49\)
\(\Rightarrow61k=49\)
\(\Rightarrow k=\dfrac{49}{61}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{49}{61}.30=\dfrac{1470}{61}\\y=\dfrac{49}{61}.16=\dfrac{784}{61}\\z=\dfrac{49}{61}.15=\dfrac{735}{61}\end{matrix}\right.\)
\(\dfrac{2x}{5}=\dfrac{3y}{2}=\dfrac{5z}{7}\)
\(\Leftrightarrow28x=105y=50z\)
hay x/75=y/20=z/42
Đặt x/75=y/20=z/42=k
=>x=75k; y=20k; z=42k
Ta có: xyz=504000
\(\Leftrightarrow k^3\cdot63000=504000\)
\(\Leftrightarrow k=2\)
=>x=150; y=40; z=84
5x/2=7z/3
nên 15x=14z
=>x/14=z/15
3x=5y nên x/5=y/3
=>x/70=y/42=z/45
Đặt x/70=y/42=z/45=k
=>x=70k; y=42k; z=45k
Tacó: xz=47250
=>3150k2=47250
=>k2=15
TH1: \(k=\sqrt{15}\)
\(x=70\sqrt{15};y=42\sqrt{15};z=45\sqrt{15}\)
TH2:
\(k=-\sqrt{15}\)
\(x=-70\sqrt{15};y=-42\sqrt{15};z=-45\sqrt{15}\)
Đặt \(x=\frac{y}{2}=\frac{z}{3}=k\left(k\in Q\right)\)\(\Rightarrow x=k;y=2k;z=3k\)
Thế (1) vào biểu thức trên
\(\Rightarrow2\left(x^2+y^2\right)-z^2=9\)
\(\Leftrightarrow2\left[\left(k\right)^2+\left(2k\right)^2\right]-\left(3k\right)^2=9\)
\(\Rightarrow2\left(k^2+4k^2\right)-9k^2=9\)
\(\Rightarrow2k^2+8k^2-9k^2=9\)
\(\Rightarrow k^2=9\)
\(\Rightarrow k=\hept{\begin{cases}3\\-3\end{cases}}\)
Với k = 3
\(\Rightarrow x=3;y=3.2=6;z=3.3=9\)
Với k = -3
\(\Rightarrow x=-3;y=-3.2=-6;z=-3.3=-9\)
\(2x=3y=4z\)
\(\Leftrightarrow\dfrac{2x}{12}=\dfrac{3y}{12}=\dfrac{4z}{12}\)
\(\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}\)
Đặt :
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=4k\\z=3k\end{matrix}\right.\)
\(2x^2-3z^2=1125\Leftrightarrow2.\left(6k\right)^2-3.\left(3k\right)^2=1125\Leftrightarrow72k^2-27k^2=1125\)
\(\Leftrightarrow45k^2=1125\)
\(\Leftrightarrow k^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
Với \(k=5\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6.5=30\\y=4.5=20\\z=3.5=15\end{matrix}\right.\)
Với \(k=-5\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6.\left(-5\right)=-30\\y=4.\left(-5\right)=-20\\z=3.\left(-5\right)=-15\end{matrix}\right.\)
Vậy ...
Bài 1:
x/-3=9/4
nên x=-9/4*3=-27/4
2x+y=-4
=>y=-4-2x=-4-2*(-27/4)=-4+27/2=27/2-8/2=19/2
Đặt x/3=y/4=z/5=k
=>x=3k; y=4k; z=5k
Ta có: \(2x^2+y^2-z^2=9\)
\(\Leftrightarrow18k^2+16k^2-25k^2=9\)
\(\Leftrightarrow9k^2=9\)
\(\Leftrightarrow k^2=1\)
TH1: k=1
=>x=3; y=4; z=5
TH2: k=-1
=>x=-3; y=-4; z=-5