1)  1/x-1/y

=y/xy-x/xy

=y-x/xy

= - (x-y)/xy

= -1 (vì x-y=xy)

2)

(x- 1/2)*(y+1/3)*(z-2)=0

=> x-1/2 = 0 hoac y+1/3=0 hoac z-2=0

th1 :x-1/2=0 => x=1/2

x+2=y+3=z+4

mà x=1/2 => y= -1/2 ; z=-3/2

th2: y+1/3=0

th3 : z-2=0

(tự làm nha)

1)  Với x,y khác 0, Ta có

\(\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}=-\left(\frac{x-y}{xy}\right)=-\left(\frac{xy}{xy}\right)=-1\)

Vậy \(\frac{1}{x}-\frac{1}{y}=-1\)

2) Ta có:

\(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\)

Trường hợp 1: x - 1/2 = 0 => x = 1/2 \(\Rightarrow\hept{\begin{cases}y=\frac{1}{2}+2-3=-\frac{1}{2}\\z=\frac{1}{2}+2-4=-\frac{3}{2}\end{cases}}\)

Trường hợp 2: y + 1/3 = 0 => y = -1/3 \(\Rightarrow\hept{\begin{cases}x=-\frac{1}{3}+3-2=\frac{2}{3}\\z=-\frac{1}{3}+3-4=-\frac{4}{3}\end{cases}}\)

Trường hợp 3: z - 2 = 0 => z = 2 \(\Rightarrow\hept{\begin{cases}x=2+4-2=4\\y=2+4-3=3\end{cases}}\)

Vậy......

CÁM ƠN NHỮNG NGƯỜI BẠN NHẤT QUẢ ĐẤT NÀY LUN

a) Áp dụng tính chất ..., ta có :

 \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+2y-z}{2+6-4}=\frac{8}{4}=2\)

\(\Rightarrow x=4;y=6;z=8\)

b)2x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{2}\)\(\Rightarrow\frac{x}{20}=\frac{y}{10}\)( 1 )

4y =5z \(\Rightarrow\frac{y}{5}=\frac{z}{4}\)\(\Rightarrow\frac{y}{10}=\frac{z}{8}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{20}=\frac{y}{10}=\frac{z}{8}\)

Áp dụng tính chất ..., ta có :

\(\frac{x}{20}=\frac{y}{10}=\frac{z}{8}=\frac{x-y+2z}{20-10+16}=\frac{40}{26}=\frac{20}{13}\)

\(\Rightarrow x=\frac{400}{13};y=\frac{200}{13};z=\frac{160}{13}\)

còn lại tương tự

  Bài 1:  \(x\).(\(x-y\)) = \(\dfrac{3}{10}\) và y(\(x-y\)) = - \(\dfrac{3}{50}\)

    \(x\)(\(x\) - y) - y(\(x\) - y) = \(\dfrac{3}{10}\) - ( - \(\dfrac{3}{50}\))

     (\(x-y\)).(\(x-y\)) = \(\dfrac{3}{10}\) + \(\dfrac{3}{50}\)

        (\(x-y\))² = \(\dfrac{15}{50}\) + \(\dfrac{3}{50}\)

        (\(x\) - y)² = \(\dfrac{9}{25}\) = (\(\dfrac{3}{5}\))²

        \(\left[{}\begin{matrix}x-y=-\dfrac{3}{5}\\x-y=\dfrac{3}{5}\end{matrix}\right.\) 

TH1 \(x-y=-\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\left(-\dfrac{3}{5}\right)=\dfrac{3}{10}\\y.\left(-\dfrac{3}{5}\right)=-\dfrac{3}{50}\end{matrix}\right.\) 

⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\left(-\dfrac{3}{5}\right)=\dfrac{-1}{2}\\y=-\dfrac{3}{50}:\left(-\dfrac{3}{5}\right)=\dfrac{1}{10}\end{matrix}\right.\) 

TH2: \(x-y=\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\dfrac{3}{5}=\dfrac{3}{10}\\y.\dfrac{3}{5}=-\dfrac{3}{50}\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\dfrac{3}{5}=\dfrac{1}{2}\\y=-\dfrac{3}{50}:\dfrac{3}{5}=-\dfrac{1}{10}\end{matrix}\right.\)  

    Vậy (\(x;y\)  ) = (- \(\dfrac{1}{2}\); \(\dfrac{1}{10}\)); (\(\dfrac{1}{2}\); - \(\dfrac{1}{10}\))

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{2x}{2.2}=\frac{3.y}{3.3}=\frac{z}{4}\Leftrightarrow\frac{\left(2x+3y+z\right)}{4+9+4}=\frac{34}{17}=2\) Tính chất tỷ lệ thức

\(\frac{x}{2}=2\Rightarrow x=2.2=4\)

\(\frac{y}{3}=2\Rightarrow y=3.2=6\)

\(\frac{z}{4}=2\Rightarrow z=4.2=8\)

a )  

Ta có : 

\(\hept{\begin{cases}\frac{x}{5}=\frac{y}{6}\\\frac{y}{8}=\frac{z}{7}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{20}=\frac{y}{24}\\\frac{y}{24}=\frac{z}{21}\end{cases}}}\)

và \(x+y-z=69\)

ADTCDTSBN , ta có : 

\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{20}=3\\\frac{y}{24}=3\\\frac{z}{21}=3\end{cases}\Rightarrow\hept{\begin{cases}x=3.20=60\\y=3.24=72\\z=3.21=63\end{cases}}}\)

Vậy ...

b )  

Ta có : 

\(5y=72\Rightarrow y=\frac{72}{5}=14,4\)

\(\Rightarrow x=14,4.3:2=21,6\)

và \(3x+5y-7z=30\)

Thay vào làm tiếp : 

c ) 

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)

\(=\frac{3\left(x-1\right)}{6}=\frac{4\left(y+3\right)}{16}=\frac{5\left(z-5\right)}{30}\)

\(=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)

\(=\frac{5z-25-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)( ADTCDTSBN ) 

\(=\frac{5z-25-3x+3-4y-12}{8}=\frac{5z-3x-4y-34}{8}\)

\(=\frac{50-34}{8}=\frac{16}{8}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=2\\\frac{y+3}{4}=2\\\frac{z-5}{6}=2\end{cases}\Rightarrow\hept{\begin{cases}x-1=2.2=4\\y+3=2.4=8\\z-5=2.6=12\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\y=5\\z=17\end{cases}}}\)

Vậy ...

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405