Ta có: \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}\) \(=\frac{x-y+z-t}{15-7+3-1}=\frac{10}{10}=1\)

=>x:15=1 =>x=15

    y:7=1=>y=7

    z:3=1=>z=3

    t:1=1=>t=1

Theo đề \(x:y:z:t=15:7:3:1\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}\)và \(x-y+z-t=10\)

Áo dụng TC dãy tỉ số bằng nhau, ta có

\(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x-y+z-t}{15-7+3-1}=\frac{10}{10}=1\)

Vậy \(x=1.15=15\)

\(y=1.7=7\)

\(z=1.3=3\)

\(t=1.1=1\)

Vậy \(\left(x;y;z;t\right)=\left(15;7;3;1\right)\)

Ta co : 

x:y:z:t=15:7:3:1 va x-y+z-t=10

Theo de bai ta co:

\(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}\) va x-y+z-t = 10

Áp dụng tính chất tỉ số bằng nhau ta có :

\(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}\Rightarrow\frac{x-y+z-t}{15-7+3-1}=\frac{10}{10}=1\)

Suy ra : \(\frac{x}{15}=1\Rightarrow x=15.1=15\)

\(\frac{y}{7}=1\Rightarrow y=1.7=7\)

\(\frac{z}{3}=1\Rightarrow z=1.3=3\)

\(\frac{t}{1}=1\Rightarrow t=1.1=1\)

Vay : x=15 ; y=7 ; z=3 ; t=1

1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)

\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)

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

=>\(x=3\cdot20=60\)

    \(y=3\cdot24=72\)

    \(z=3\cdot21=63\)

3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)

=> \(x=1\cdot15=15\)

     \(y=1\cdot7=7\)

     \(z=1\cdot3=3\)

     \(t=1\cdot1=1\)

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

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{t}{1}=\dfrac{x-y+z-t}{15-7+3-1}=\dfrac{10}{10}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=15\\y=7\\z=3\\t=1\end{matrix}\right.\)

15 - 7 + 3 + 1 = 12 

\(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x-y+z-t}{15-7+3-1}=\frac{10}{10}=1\)

x=15

y=7

z=3

t=1

z = 3

Tick mk nha  _Công chúa nhỏ _

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

Ta có:

x5=y6⇒x20=y24x5=y6⇒x20=y24   (1)(1)

y8=z7=y24=z21y8=z7=y24=z21    (2)(2)

Từ (1)(1) và (2)(2) ⇒x20=y24=z21⇒x20=y24=z21

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

x20=y24=z21=x+y−z20+24−21=6923=3x20=y24=z21=x+y-z20+24-21=6923=3

⇒⎧⎪⎨⎪⎩x=60y=72z=63⇒{x=60y=72z=63

Vậy x=60;y=72x=60;y=72 và z=63

a, \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}vàx+y-10=26\)

=> x + y - 10 = 26

x + y = 26 + 10 = 36

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{36}{7}\)

Suy ra:

x = \(\dfrac{36.3}{7}=\dfrac{108}{7}\)

\(y=\dfrac{36.4}{7}=\dfrac{144}{7}\)

\(z=\dfrac{36.5}{7}=\dfrac{180}{7}\)

Vậy\(x=\dfrac{108}{7},y=\dfrac{144}{7},z=\dfrac{180}{7}.\)