1, \(4x=5y\) 

mà  \(y-2x=-5\)

\(\Rightarrow x=\frac{y+5}{2}\)

\(\Rightarrow\left(\frac{y+5}{2}\right).4=5y\)

\(\Rightarrow\frac{4y+20}{2}=5y\)

\(\Rightarrow2y+10=5y\)

\(\Rightarrow10=3y\)

\(\Rightarrow y=\frac{10}{3}\)

\(\Rightarrow x=\frac{y+5}{2}=\frac{\frac{10}{3}+5}{2}=\frac{\frac{25}{3}}{2}=\frac{25}{6}\)

Vậy \(x=\frac{25}{6};y=\frac{10}{3}\)

b, \(\frac{x}{3}=\frac{y}{4}\)

mà \(xy=192\)

Gọi \(x=3k\)

        \(y=4k\)

\(\Rightarrow3k.4k=192\)

\(\Rightarrow12.k^2=192\)

\(\Rightarrow k^2=\frac{192}{12}\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k^2=4^2\)

\(\Rightarrow k=4\)

\(\Rightarrow x=3k=3.4=12\)

\(\Rightarrow y=4k=4.4=16\)

Vậy \(x=12;y=16\)

Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=k\)

Theo đề bài, ta có :

\(xy=54\Rightarrow2k.3k=54\)

\(\Rightarrow5k=54\Rightarrow k=10,8\)

Ta thấy :

\(\dfrac{x}{2}=10,8\Rightarrow x=10,8.2=21,6\)

\(\dfrac{y}{3}=10,8\Rightarrow y=10,8.3=32,4\)

Đặt :\(\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

mà \(xy=54\)

hay 2k . 3k = 54

\(\Rightarrow6.k^2=54\)

\(\Rightarrow k^2=9=\left(\pm3\right)^2\)

Với k = 3 \(\Rightarrow\) \(x=2.3=6;y=3.3=9\)

Với k = -3 \(\Rightarrow x=2.\left(-3\right)=-6;y=3.\left(-3\right)=-9\)

Đặt x = 4k

y = 7k

=> 4k.7k = 112

=> 28.k^2 = 112

=> k^2 = 112 : 28 = 4

=> k = 2

=> x = 4.2 = 8

y = 7.2 = 14

14 nha

\(\frac{2}{x}=\frac{3}{y}\Rightarrow x=\frac{2y}{3}\)

Thay vào x . y, ta được:

 \(x\cdot y=\frac{2y}{3}\cdot y=\frac{2y^2}{3}=96\)

=>  \(2y^2=96\cdot3=288\Rightarrow y^2=\frac{288}{2}=144\)

=>  \(y=\sqrt{144}=12\)  hoặc   \(y=-12\)

• y = 12               =>   x = 96 : 12 = 8
• y = -12              =>  x = 96 : (-12) = -8

Vậy x = 8; y = 12    hoặc    x = -8 ; y = -12  

\(\frac{2}{x}=\frac{3}{y}=>\frac{2}{x}.\frac{3}{y}=\frac{3}{y}.\frac{3}{y}=>\frac{6}{xy}=\frac{9}{y^2}=>\frac{6}{96}=\frac{9}{y^2}=>\frac{1}{16}=\frac{9}{y^2}\)

\(=>y^2=9:\frac{1}{16}=144=12^2=\left(-12\right)^2\)

=>y=12,-12

Với y=12=>x=96:12=8

Với y=-12=>x=96:(-12)=-8

Vậy x=-8,y=-12

       x=8,y=12

x/2 = y/5

=> xy/10 = x/2 = y/5 = 10/10 = 1

=> x = 1x 2 = 2

     y = 1 x 5 = 5

Đặt \(k=\frac{x}{2}=\frac{y}{5}\)

=> \(k^2=\frac{xy}{2.5}=\frac{xy}{10}=\frac{10}{100}=1\)

=> k = -1;1

+ k = -1 thì \(\frac{x}{2}=-1\Rightarrow x=-2\)

                 \(\frac{y}{5}=-1\Rightarrow y=-5\)

+ k = 1 thi \(\frac{x}{2}=1\Rightarrow x=2\)

                 \(\frac{y}{5}=1\Rightarrow y=5\) 

Vậy .............................

Ta có : x + y = 3 => x = 3 - y 

=> \(xy=\left(3-y\right)y=3y-y^2=-\left(y^2-3y\right)=-\left[y^2-2.y.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right]\)

\(=-\left[\left(y-\frac{3}{2}\right)^2-\frac{9}{4}\right]=-\left(y-\frac{3}{2}\right)^2+\frac{9}{4}\)

Vì \(-\left(y-\frac{3}{2}\right)^2\le0\) \(\forall x\)

\(\Rightarrow-\left(y-\frac{3}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\) \(\forall x\)

Dấu "=" xảy ra <=> \(-\left(y-\frac{3}{2}\right)^2=0\Rightarrow y=\frac{3}{2}\Rightarrow x=3-\frac{3}{2}=\frac{3}{2}\)

Vậy GTNN của xy là \(\frac{9}{4}\) tại \(x=y=\frac{3}{2}\)

GTNN của xy là 9/4 tại x = y = 3/2

   Mà bạn Đinh Đức Hùng có hack không vậy? Sao bạn ấy nhiều điểm thế! (không có ý nói xấu bạn đâu nha! Đừng hiểu lầm mình)

1. Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)

+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)

+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)

+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)

Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)

2,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)

+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)

+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)

+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)

Vậy \(x=-2;y=-3;c=-4\)

Đặt \(\frac{x}{3}=\frac{y}{9}=k\Rightarrow x=3k;y=9k\)

\(\Rightarrow xy=3k.9k=27k^2=675\)

\(\Rightarrow k^2=25\Rightarrow k=\pm5\)

\(\Rightarrow x=3.\pm5=\pm15\)

     \(y=9.\pm5=\pm45\)

Vậy ....

( \(\pm\) là âm hoặc dương nha)

Đặt \(\frac{x}{3}=\frac{y}{9}=k\)

\(\Rightarrow x=3k;y=9k\)

\(\Rightarrow xy=3k.9k=27k^2=675\)

\(\Rightarrow k^2=25\Rightarrow k=5;-5\)

Với k = 5=> x = 15 ; y = 45

Với k = -5=> x = -15 ; y = -45