Từ 3 pt dễ dàng suy ra x;y;z đều không âm

Do đó: \(12x^2=y\left(9x^2+4\right)\ge y.2\sqrt{9x^2.4}=12xy\Rightarrow x\ge y\)

Tương tự: \(12y^2=z\left(9y^2+4\right)\Rightarrow y\ge z\)

\(12z^2=x\left(9z^2+4\right)\Rightarrow z\ge x\)

\(\Rightarrow x=y=z\)

Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}x=y=z=0\\x=y=z=\frac{3}{2}\end{matrix}\right.\)

a) \(-9x^2+12x-15=-\left(9x^2-12x+4\right)-11=-\left(3x-2\right)^2-11\le11< 0\)

b) \(-2x^2+4x-9=-2\left(x^2-2x+1\right)-7=-2\left(x-1\right)^2-7\le-7< 0\)

c) \(xy-x^2-y^2-1=-\dfrac{1}{2}\left(2x^2+2y^2-2xy+2\right)=-\dfrac{1}{2}\left[\left(x-y\right)^2+x^2+y^2+2\right]< 0\)

\(3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{15}=\dfrac{y}{9};9z=7y\Rightarrow\dfrac{z}{7}=\dfrac{y}{9}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}\)

Áp dụng...

\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}=\dfrac{3x}{45}=\dfrac{2y}{18}=\dfrac{4z}{28}=\dfrac{3x-2y-4z}{45-18-28}=\dfrac{10}{-1}=-10\\ \Rightarrow\left\{{}\begin{matrix}x=-150\\y=-90\\z=-70\end{matrix}\right.\)

\(3x=5y\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}\)

hay \(\dfrac{x}{15}=\dfrac{y}{9}\left(1\right)\)

7y=9z

nên \(\dfrac{y}{9}=\dfrac{z}{7}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{4}=\dfrac{3x-2y-4z}{45-18-16}=\dfrac{10}{11}\)

Do đó: \(x=\dfrac{150}{11};y=\dfrac{90}{11};z=\dfrac{40}{11}\)

\(3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{15}=\dfrac{y}{9}\)

\(9z=7y\Rightarrow\dfrac{y}{9}=\dfrac{z}{7}\)

Áp dung t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}=\dfrac{3x}{45}=\dfrac{2y}{18}=\dfrac{4z}{28}=\dfrac{3x-2y-4z}{45-18-28}=\dfrac{10}{-1}=-10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).15=-150\\y=\left(-10\right).9=-90\\z=\left(-10\right).7=-70\end{matrix}\right.\)

\(3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{15}=\dfrac{y}{9}\)

\(9z=7y\Rightarrow\dfrac{y}{9}=\dfrac{z}{7}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{15}=\dfrac{y}{9}=\dfrac{z}{7}=\dfrac{3x-2y-4z}{45-18-28}=\dfrac{10}{-1}=-10\)

\(\dfrac{x}{15}=-10\Rightarrow x=-150\\ \dfrac{y}{9}=-10\Rightarrow y=-90\\ \dfrac{z}{7}=-10\Rightarrow z=-70\)

Ta có:  

Khi đó:  

Suy ra (Q): 2y+3z-11=0

Ta có: 9x=12y=4z => \(\frac{9x}{36}\)=\(\frac{12y}{36}\)=\(\frac{4z}{36}\)  => \(\frac{x}{4}\)= \(\frac{y}{3}\)=\(\frac{z}{9}\) => \(\frac{x}{4}\)=\(\frac{3y}{9}\)=\(\frac{4z}{36}\)

và x-3y+4z=62.

Áp dụng t/c của dãy tỉ số bằng nhau, ta có: \(\frac{x}{4}\)=\(\frac{3y}{9}\)=\(\frac{4z}{36}\)= \(\frac{x-3y+4z}{4-9+36}\)= \(\frac{62}{31}\)= 2

Do đó:

x=2.4=8

3y=2.9=18 => y=6

4z=2.36=72 => z=18.

Vậy x=8, y=6, z=18

~Hok tốt!~

Theo bài cho , ta có :

\(9x=12y=4z\)

\(\Rightarrow\frac{9x}{36}=\frac{12y}{36}=\frac{4z}{36}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{3}=\frac{z}{9}\)

\(\Rightarrow\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}\)   và \(x-3y+4z=62\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}=\frac{x-3y+4z}{4-9+36}=\frac{62}{31}=2\)

\(+)\frac{x}{4}=2\Rightarrow x=8\)

\(+)\frac{3y}{9}=2\Rightarrow3y=18\Rightarrow y=6\)

\(+)\frac{4z}{36}=2\Rightarrow4z=72\Rightarrow z=18\)

Vậy x = 8 , y = 6 và z = 18 .

Học tốt

\(\hept{\begin{cases}9x=12y=4z\\x-3y+4z=62\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{9}}=\frac{y}{\frac{1}{12}}=\frac{z}{\frac{1}{4}}\\x-3y+4z=62\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{9}}=\frac{3y}{\frac{1}{4}}=\frac{4z}{1}\\x-3y+4z=62\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{\frac{1}{9}}=\frac{3y}{\frac{1}{4}}=\frac{4z}{1}=\frac{x-3y+4z}{\frac{1}{9}-\frac{1}{4}+1}=\frac{62}{\frac{31}{36}}=72\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{9}}=72\Rightarrow x=8\\\frac{y}{\frac{1}{12}}=72\Rightarrow y=6\\\frac{z}{\frac{1}{4}}=72\Rightarrow z=18\end{cases}}\)

Vậy x = 8 ; y = 6 ; z = 18

a) \(2x=5y\)⇒\(x=\dfrac{5}{2}y\)⇒\(xy=\dfrac{5}{2}y^2\)

Thay \(xy=250\), ta có:

\(250=\dfrac{5}{2}y^2\)

⇒\(y^2=100\)⇒\(y=+-10\)

+) \(y=10\text{⇒}x=250:10=25\)

+) \(y=-10\text{⇒}x=250:-10=-25\)

\(a,2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=k\\ \Rightarrow x=5k;y=2k\\ xy=250\Rightarrow5k\cdot2k=250\Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=25;y=10\\x=-25;y=-10\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{4}=a\Rightarrow x=3a;y=2a;z=4a\\ xyz=192\Rightarrow24a^3=192\Rightarrow a^3=8\Rightarrow a=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=4\\z=8\end{matrix}\right.\\ c,\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=q\Rightarrow x=5q;y=2q;z=-3q\\ xyz=240\Rightarrow-30q^3=240\Rightarrow q^3=-8\Rightarrow q=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-10\\y=-4\\z=6\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}2x=5y\\xy=250\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5y=0\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2xy-5y^2=0\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y^2=500\\2xy=500\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10\\x=25\end{matrix}\right.\)