Có \(\frac{3x-y}{x+y}=\frac{1}{2}\)

=> (3x- y).2 = x + y

=> 6x - 2y = x + y 

=> 5x = 3y

=> \(\frac{x}{y}=\frac{3}{5}\)

\(\frac{3x-y}{x+y}=\frac{1}{2}\)

\(\Rightarrow2\left(3x-y\right)=x+y\)

\(\Rightarrow6x-2y=x+y\)

\(\Rightarrow5x=3y\)

\(\Rightarrow\frac{x}{y}=\frac{3}{5}\)

Vậy :................

\(\frac{3x-y}{x+y}=\frac{1}{2}=>\left(3x-y\right).2=x+y=>6x-2y=x+y=>6x-x=2y+y\)

=>5x=3y

hay x/y=3/5

vậy tỉ số x/y=3/5

\(\frac{3x-y}{x+y}=\frac{1}{2}\)

<=> 2(3x - y) = x + y

=> 6x - 2y = x + y

=> 5x = 3y

=> \(\frac{x}{y}=\frac{3}{5}\)

4(3x-y) = 3(x+y)

12x -4y = 3x+3y

9x = 7y

x/y = 7/9

Đặt: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

\(\frac{4x-3y+z}{3x+y-2z}=\frac{4.2k-3.3k+5k}{3.2k+3k-2.5k}=\frac{4k}{-1k}=-4\)

Ta có: \(\frac{3x-y}{x+y}\)=\(\frac{3}{4}\)

\(\Leftrightarrow\)4(3x-y)=3(x+y)

\(\Leftrightarrow\)12x-4y=3x+3y

\(\Leftrightarrow\)12x-3x=4x+3y

\(\Leftrightarrow\)9x=7y

\(\Leftrightarrow\)\(\frac{x}{y}\)=\(\frac{7}{9}\)

\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow\frac{3x-y}{x+y}+1=\frac{3}{4}+1\Leftrightarrow\frac{4x}{x+y}=\frac{7}{4}.\) Ở vế trái chia cả tử và mẫu cho y , được:

\(\frac{4.\frac{x}{y}}{\frac{x}{y}+1}=\frac{7}{4}\)  Suy ra :  \(16.\frac{x}{y}=7\left(\frac{x}{y}+1\right)\) Vậy \(9.\frac{x}{y}=7\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)

Ta có: \(\frac{3x-y}{x+y}=\frac{3}{4}\)

      \(\Rightarrow\frac{3x+3y-4y}{x+y}=\frac{3}{4}\)

      \(\Rightarrow3-\frac{4y}{x+y}=\frac{3}{4}\)

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

       \(\Rightarrow4.4y=9.\left(x+y\right)\)

       \(\Rightarrow16y=9y+9x\)

       \(\Rightarrow9x=16y-9y=7y\)

       \(\Rightarrow\frac{x}{y}=\frac{7}{9}\)

Vậy tỉ số \(\frac{x}{y}=\frac{7}{9}\)

này thì mk chịu

Câu a:

0,4 : x = x : 0,9

x^2 = 0,4 x 0,9

x^2 = 0,36

x^2 = (0,6)^2

x = - 0,6 hoặc x = 0,6

Vậy x ∈ {-0,6; 0,6}

Câu b:

13\(\frac13\) : 1\(\frac13\) = 26 : (2x - 1)

\(\frac{40}{13}\) : \(\frac43\) = 26 :(2x - 1)

\(\frac{40}{13}\) x \(\frac34\) = 26 :(2x - 1)

\(\frac{30}{13}\) = 26 : (2x - 1)

26 : \(\frac{30}{13}\) = 2x - 1

26 x \(\frac{13}{30}\) = 2x - 1

169/15 = 2x - 1

2x = 169/15 + 1

2x = 184/15

x = 184/15 : 2

x = 184/15 x 1/2

x = 92/15

Vậy x = 92/15

Câu 1:

\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)

Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)

\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)

Dấu = xảy ra khi x=y=1/2

Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)

CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu = xảy ra khi x=y=z=1