\(2x=3y=4z\)

\(\Leftrightarrow\dfrac{2x}{12}=\dfrac{3y}{12}=\dfrac{4z}{12}\)

\(\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}\)

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 ...

google xin tài trợ chương trình

có google thôi anh

\(\hept{\begin{cases}x^2-2x\sqrt{y}+2y=x\\y^2-2y\sqrt{z}+2z=y\\z^2-2z\sqrt{x}+2x=z\end{cases}}\)

\(\Leftrightarrow x^2-2x\sqrt{y}+2y+y^2-2y\sqrt{z}+2z+z^2-2z\sqrt{x}+2x=x+y+z\)

\(\Leftrightarrow\left(x-\sqrt{y}\right)^2+\left(y-\sqrt{z}\right)^2+\left(z-\sqrt{x}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\sqrt{y}=0\\y-\sqrt{z}=0\\z-\sqrt{x}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{y}\\y=\sqrt{z}\\z=\sqrt{x}\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=y=z=0\\x=y=z=1\end{cases}}\)

Bó tay. com

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\(\frac{x}{10}=\frac{y}{5}\Rightarrow\frac{x}{20}=\frac{y}{10}\)

\(\frac{y}{2}=\frac{z}{3}\Rightarrow\frac{y}{10}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{20}=\frac{y}{10}=\frac{z}{15}\)

\(\Rightarrow\frac{2x}{40}=\frac{3y}{30}=\frac{z}{15}\)

áp dụng tc của dãy tỉ số = nhau

=> x/20=y/10; y/10=z/15

=> x/20=y/10=z/15

từ...áp dụng....

đc : (2x-3y+4z)/(40-30+60)=280/70=4

  => x=..

=> y=...

=> z=...

bạn tự làm nha