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

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{-4}=\frac{3x}{3.2}=\frac{2z}{2.\left(-4\right)}=\frac{3x-2z}{6-\left(-8\right)}=\frac{28}{14}=2\)

\(\hept{\begin{cases}\frac{x}{2}=2\Rightarrow x=2.2=4\\\frac{y}{3}=2\Rightarrow y=2.3=6\\\frac{z}{-4}=2\Rightarrow z=-4.2=-8\end{cases}}\)

Vậy x=4,y=6,z=-8

a. Theo t/c dãy tỉ số = nhau:

\(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{42}{7}=6\)

=>\(\frac{x}{2}=6\Rightarrow x=6.2=12\)

=>\(\frac{y}{5}=6\Rightarrow y=6.5=30\)

Vậy x=12; y=30.

b. \(\left|x-0,25\right|-\frac{5}{6}=1\frac{2}{3}\)

=> \(\left|x-0,25\right|=1\frac{2}{3}+\frac{5}{6}\)

=> \(\left|x-0,25\right|=\frac{5}{2}=2,5\)

+) x-0,25=2,5

=> x=2,5+0,25

=> x=2,75

+) x-0,25=-2,5

=> x=-2,5+0,25

=> x=-2,25

Vậy x \(\in\){-2,25; 2,75}.

c. y=kx

=> -17=k.8

=> k=-17/8

Vậy hệ số tỉ lệ là -17/8.

a) \(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{42}{7}=6\)

=> x=12   ;   y = 30

b)  \(\left|x-0,25\right|-\frac{5}{6}=1\frac{2}{3}=>\left|x-0,25\right|=\frac{5}{3}+\frac{5}{6}=\frac{5}{2}=2,5\)

=> x-0,25 = 2,5    hoac:  -2,5

=> x = 2,75      hoac x= -2,25

Vay: x la { 2,75  ;   -2,25 }

c) Ti le gi vay ban.

Neu thuan thi he so ti le la: \(-\frac{17}{8}\)

Neu nghich thi he so ti le la : -136

a) \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{x+y+z}{10+6+21}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

-> \(5x=2\cdot50=100\) => \(x=\frac{100}{5}=20\)

\(y=2\cdot6=12\)

\(2z=2\cdot42=84\) => \(z=\frac{84}{2}=42\)

Bài 1: Tìm x, y, z

\(\frac{x}{3}=\frac{y}{4}=>\frac{x}{3\times3}=\frac{y}{4\times3}=>\frac{x}{9}=\frac{y}{12}\)

\(\frac{y}{3}=\frac{z}{5}=>\frac{y}{3.4}=\frac{z}{5.4}=>\frac{y}{12}=\frac{z}{20}\)

=> \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

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

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\) -> \(\frac{2x}{2\times9}=\frac{3y}{3\times12}=\frac{z}{20}\) -> \(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)

-> \(\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

\(\frac{x}{9}=3\rightarrow x=27\)

\(\frac{y}{12}=3\rightarrow y=36\)

\(\frac{z}{20}=3\rightarrow z=60\)

Vậy x = 27 ; y = 36 ; z = 60

Bài 2 : Tìm x, y:

5x = 2y và x.y = 40

Vì 5x = 2y => \(\frac{x}{2}=\frac{y}{5}\)

Cách 1:

\(\frac{x}{2}=\frac{y}{5}\) và x.y = 40

Đặt \(\frac{x}{2}=\frac{y}{5}\) = k

=> x = 2.k ; y = 5.k

x.y = 40 -> 2k = 5k = 40

-> 10 . \(k^2\) = 40

-> \(k^2\) = 4 -> k = 2 hoặc k = -2

k = 4 ta có : \(\frac{x}{2}=\frac{y}{5}=2->x=4;y=10\)

k = -4 ta có : \(\frac{x}{2}=\frac{y}{5}=-2->x=-4;y=-10\)

Cách 2:

\(\frac{x}{2}=\frac{y}{5}->\frac{x.x}{2}=\frac{x.y}{5}->\frac{x^2}{2}=\frac{40}{5}=\frac{x^2}{2}=8\)

=> \(x^2\) = 8 . 2 = 16 -> x = 4 hoặc -4

x = 4 -> 4.y = 40 => y = 10

x = -4 -> (-4).y = 40 => y = -10

Vậy x = 4 hoặc -4

y = 10 hoặc -10

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\\\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}=\frac{2x}{18}=\frac{-3y}{-36}=\frac{z}{15}=\frac{2x-3y+z}{18-\left(-36\right)+15}=\frac{6}{69}=\frac{2}{23}\)Suy ra x =\(\frac{2}{23}\cdot9=\frac{18}{23}\)

\(y=\frac{2}{23}\cdot12=\frac{24}{23}\\ z=\frac{2}{23}.15=\frac{30}{23}\)

Ta có :

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy ...

Ta có: \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\frac{x+4}{5}=\frac{20}{x+4}\)

\(\Leftrightarrow\left(x+4\right)^2=100\)

\(\Leftrightarrow x+4=\pm\sqrt{100}\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)

x=6  nha