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a) \(\dfrac{3cy-4bz}{2x}=\dfrac{4az-2cx}{3y}=\dfrac{2bx-3ay}{4z}\)
=> \(\dfrac{3cy-4bz}{2x}.\dfrac{2x}{2x}=\dfrac{4az-2cx}{3y}.\dfrac{3y}{3y}=\dfrac{2bx-3ay}{4z}.\dfrac{4z}{4z}\)
=> \(\dfrac{6cxy-8bzx}{4x^2}=\dfrac{12azy-6cxy}{9y^2}=\dfrac{8bxz-12ayz}{16z^2}\)
Áp dụng t/c ...
\(\dfrac{6cxy-8bzx}{4x^2}=\dfrac{12azy-6cxy}{9y^2}=\dfrac{8bxz-12ayz}{16z^2}=\dfrac{6cxy-8bzx+12azy-6cxy+8bxz-12ayz}{4x^2+9y^2+16z^2}=\dfrac{0}{4x^2+9y^2+16z^2}=0\)
Ta có : 6cxy - 8bzx = 0
=> 6cxy = 8bzx
=>3cx = 4bz
=>\(\dfrac{c}{4z}=\dfrac{b}{3y}\) (1)
Ta có : 12azy - 6cxy = 0
=> 12azy = 6cxy
=> 4az = 2cx
=> \(\dfrac{a}{2x}=\dfrac{c}{4z}\) (2)
Từ (1),(2) => \(\dfrac{a}{2x}=\dfrac{b}{3y}=\dfrac{c}{4z}\) (ĐPCM)
\(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\) va \(x+y-z=69\)
Ta co: \(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\) ; \(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)
➤ \(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\) ➤ \(\dfrac{x+y-z}{20+24-21}\)
➤ \(\dfrac{69}{23}=3\) ➤ \(x=20.3=60\)
\(y=24.3=72\)
\(z=21.3=63\)
\(Vay\) \(x=60;y=72;z=63\)
\(2a=3b;5b=7c\) va \(3a+5c-7c=30\)
Ta co: \(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\)
\(5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\)
⇒ \(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\) ⇒ \(\dfrac{3a}{63}=\dfrac{5c}{50}=\dfrac{7b}{98}\) ⇒ \(\dfrac{3a+5c-7b}{63+50-98}\)
⇒ \(\dfrac{30}{15}=2\) ➤ \(3a=63.2=126\) ➤ \(a=126:3=42\)
\(5c=50.2=100\) \(c=100:5=20\)
\(7b=98.2=196\) \(b=196:7=28\)
Vay \(a=42;c=20;b=28\)
\(x\div y\div z=3\div8\div5\) va \(3x+y-2z=14\)
Ta co: \(x\div y\div z=3\div8\div5\Rightarrow\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\)
⇒ \(\dfrac{3x}{9}=\dfrac{y}{8}=\dfrac{2z}{10}\) ⇒ \(\dfrac{3x+y-2z}{9+8-10}\)
⇒ \(\dfrac{14}{7}=2\) ➤ \(3x=9.2=18\) ➤ \(x=18:3=6\)
\(y=8.2\) \(y=16\)
\(2z=10.2=20\) \(z=20:2=10\)
Vay \(x=6;y=16;z=10\)
Chuc ban hoc tot 
TH1: \(x+y+z+t\ne0\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)
TH2: \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)
Từ gt của đề bài :
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{y+z+t}\text{=}\dfrac{y}{z+t+x}\text{=}\dfrac{z}{x+y+t}\text{=}\dfrac{t}{x+y+z}\text{=}\dfrac{x+y+z+t}{3.\left(x+y+z+t\right)}\left(\cdot\right)\)
Xét TH : \(x+y+z+t\text{=}0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z\text{=}-\left(x+t\right)\\z+t\text{=}-\left(x+y\right)\\x+t\text{=}-\left(y+z\right)\end{matrix}\right.\)
Do đó : \(P\text{=}-1+-1+-1+-1\)
\(P\text{=}-4\in Z\)
TH : \(x+y+z+t\ne0\)
\(\Rightarrow\left(\cdot\right)\text{=}\dfrac{1}{3}\)
Do đó : \(\dfrac{x}{y+z+t}\text{=}\dfrac{1}{3}\Rightarrow3x\text{=}y+z+t\)
\(\Rightarrow4x\text{=}x+y+z+t\)
\(CMTT:\left\{{}\begin{matrix}4y\text{=}x+y+z+t\\4z\text{=}x+y+z+t\\4t\text{=}x+y+z+t\end{matrix}\right.\)
Mà : \(\dfrac{x}{y+z+t}\text{=}\dfrac{y}{x+z+t}\text{=}\dfrac{z}{x+y+t}\text{=}\dfrac{t}{x+y+z}\)
\(\Rightarrow4x\text{=}4y\text{=}4z\text{=}4t\)
\(\Rightarrow x\text{=}y\text{=}z\text{=}t\)
Do đó : \(P\text{=}4\in Z\)
\(\Rightarrowđpcm\)
Kham khảo :
https://olm.vn/cau-hoi/cho-cac-so-thuc-xyzt-thoa-mandfracxyztdfracyztxdfracztxydfractxyz-cmr-p-dfracxyztdfracyztx.8377111224063.
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