1. Theo t/c của dãy tỉ số bằng nhau ta có : 

\(\frac{x}{2}=\frac{y}{5}=\frac{x.y}{2.5}=\frac{90}{10}=9\)

\(\frac{x}{2}=9\Rightarrow x=9.2=18\)

\(\frac{y}{5}=9\Rightarrow y=9.5=45\)

Vậy x = 18 ; y = 45 

sai rùi

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 :................

x+y+z=1?

Giải:

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

\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+x+z+1+x+y-2}=\frac{x+y+z}{2x+2y+2z}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow x+y+z=1\)

Xét \(\frac{x}{y+z+1}=\frac{1}{2}\)

\(\Rightarrow2x=y+z+1\)

\(\Rightarrow3x=x+y+z+1\)

\(\Rightarrow3x=1+1\)

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

Xét \(\frac{y}{x+z+1}=\frac{1}{2}\)

\(\Rightarrow2y=x+z+1\)

\(\Rightarrow3y=x+y+z+1\)

\(\Rightarrow3y=1+1\)

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

Xét \(\frac{z}{x+y-2}=\frac{1}{2}\)

\(\Rightarrow2z=x+y-2\)

\(\Rightarrow3z=x+y+z-2\)

\(\Rightarrow3z=1-2\)

\(\Rightarrow z=\frac{-1}{3}\)

Từ đó \(\frac{x+y}{z+1}=\frac{\frac{2}{3}+\frac{2}{3}}{\frac{-1}{3}+1}=\frac{\frac{4}{3}}{\frac{2}{3}}=\frac{4}{2}=2\)

Vậy \(\frac{x+y}{z+1}=2\)

Cảm ơn cậu ạ =))))))

D nha bạn

thanks

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

\(\Rightarrow12x+20y=x-2y\Leftrightarrow11x=-22y\Rightarrow\frac{x}{y}=-\frac{22}{11}=-2\)

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

<=>12x-4y=3x+3y

<=>9x=7y

<=>x/y=7/9

anh nha

đăng nghịch hay đăng thật đó bài dễ như vậy mà ko biết làm à
 

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

= \(\frac{x+y+z}{2x+2y+2z}\)= \(\frac{x+y+z}{2\left(x+y+z\right)}\)= \(\frac{1}{2}\)

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

=> \(\frac{z+1}{x+y}\)= \(\frac{1}{2}\)=> \(\frac{x+y}{z+1}\)= 2