\(\dfrac{x}{3}=\dfrac{y}{6}=\dfrac{2x^2}{18}=\dfrac{y^2}{36}=\dfrac{2x^2-y^2}{18-36}=\dfrac{-8}{-18}=\dfrac{4}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4.3}{9}=\dfrac{4}{3}\\y=\dfrac{4.6}{9}=\dfrac{8}{3}\end{matrix}\right.\)

\(x+y+z+8=2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\left(1\right)\)

Áp dụng Bđt Bunhiacopxki :

\(\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le\left(2^2+4^2+6^2\right)\left(x-1+y-2+z-3\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z-6\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z+8\right)-784\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=\dfrac{x+y+z-6}{14}\left(2\right)\)

Đặt \(t=x+y+z+8\)

\(\left(1\right)\Leftrightarrow t^2=56t-784\)

\(\Leftrightarrow t^2-56t+784=0\)

\(\Leftrightarrow\left(t-28\right)^2=0\)

\(\Leftrightarrow t=28\)

\(\Leftrightarrow x+y+z+8=28\)

\(\Leftrightarrow x+y+z-6=14\)

\(\left(2\right)\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.4=4\\z-2=1.8=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=6\\z=10\end{matrix}\right.\) thỏa mãn đề bài

Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\)

nên \(\dfrac{x}{7}=\dfrac{y}{20}\)(1)

Ta có: \(\dfrac{y}{z}=\dfrac{5}{8}\)

nên \(\dfrac{y}{5}=\dfrac{z}{8}\)

hay \(\dfrac{y}{20}=\dfrac{z}{32}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

hay \(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)

mà 2x-5y+2z=100

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

\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x-5y+2z}{14-100+64}=\dfrac{100}{-22}=\dfrac{-50}{11}\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{7}=\dfrac{-50}{11}\\\dfrac{y}{20}=\dfrac{-50}{11}\\\dfrac{z}{32}=-\dfrac{50}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{350}{11}\\y=\dfrac{-1000}{11}\\z=\dfrac{-1600}{11}\end{matrix}\right.\)

Ta có:  \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\Rightarrow\dfrac{x}{14}=\dfrac{y}{40}\Rightarrow\dfrac{2x}{28}=\dfrac{5y}{200}\) \(\left(1\right)\)

Lại có:  \(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{40}=\dfrac{z}{64}\Rightarrow\dfrac{5y}{200}=\dfrac{2z}{128}\)   \(\left(2\right)\)

Kết hợp ( 1 ) và ( 2 ) ta có:     \(\dfrac{2x+5y-2z}{28+200-128}=\dfrac{100}{100}=1\)

⇒  \(\dfrac{2x}{28}=1\Rightarrow x=\dfrac{1.28}{2}=14\)

⇒  \(\dfrac{5y}{200}=1\Rightarrow y=\dfrac{1.200}{5}=40\)

⇒  \(\dfrac{2z}{128}=1\Rightarrow z=\dfrac{1.128}{2}=64\)

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Ta có:

\(xy=x:y\Leftrightarrow xy=x.\dfrac{1}{y}\)

\(\Leftrightarrow xy-x.\dfrac{1}{y}=0\)

\(\Leftrightarrow x\left(y-\dfrac{1}{y}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\)

TH1: \(x=0\)

\(\Rightarrow x-y=xy=0\Leftrightarrow x=y=0\left(ktm\right)\)

TH2:\(y-\dfrac{1}{y}=0\Leftrightarrow\dfrac{y^2-1}{y}=0\)

\(\Leftrightarrow y^2-1=0\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)

Khi \(y=1\) thì \(x-1=x\)(không có \(x\) thoả mãn)

Khi \(y=-1\) thì \(x+1=-x\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)(tm)

Vậy \(x=-\dfrac{1}{2}\) và \(y=-1\)

z đâu bạn?

\(\dfrac{x}{y}=\dfrac{2}{5}=\dfrac{x}{2}=\dfrac{y}{5}\)

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=k2\\y=k5\end{matrix}\right.\)

mà \(xy=40\)

\(\Rightarrow2k.5k=40\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{5}=4\\\dfrac{x}{2}=\dfrac{y}{5}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=8;y=20\\x=-8;y=-20\end{matrix}\right.\)

từ x+y=xy

=>x=xy-y=y(x-1)

mà x+y=x/y

=>x+y=x-1

=>x+y=x+(-1)

=>y=-1

thay y=-1 vào x+y=xy

=>x-1=-x

=>2x=1=>x=1/2

vậy x=1/2;y=-1

tick nhé

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{24}{2}=12\)

=> \(\left\{{}\begin{matrix}x=36\\y=60\end{matrix}\right.\)

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

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{24}{2}=12 \\ \Rightarrow x=12.3=36 \\ y=12.5=60\)

Vậy...

Ta có: x:3=y:5

nên \(\dfrac{x}{3}=\dfrac{y}{5}\)

mà y-x=24

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{24}{2}=12\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{3}=12\\\dfrac{y}{5}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=36\\y=60\end{matrix}\right.\)

Vậy: (x,y)=(36;60)

a, Xét \(\dfrac{x}{-5}=2\Rightarrow x=-10\)

\(\dfrac{y}{4}=2\Leftrightarrow y=8\)

b, \(xy=6\Rightarrow x;y\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)