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 =\(\frac{2}{23}\cdot9=\frac{18}{23}\)

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

\(1.\)

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

\(\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\) \(\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\) \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

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

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

\(\frac{x}{9}=3\Rightarrow x=3.9=27\)

\(\frac{y}{12}=3\Rightarrow y=3.12=36\)

\(\frac{z}{20}=3\Rightarrow z=3.20=60\)

Vậy x = 27; y = 36 và z = 60

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)}{2.2}=\frac{3\left(y-2\right)}{3.3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=k\)

Áp dụng TC DTSBN ta có :

\(k=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}=\frac{\left(2x+3y-z\right)-5}{9}=\frac{50-5}{9}=5\)

\(\Rightarrow x-1=10;y-2=15;z-3=20\)

\(\Rightarrow x=11;y=17;z=23\)

ta có \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}\)và x.y=48

xét \(\frac{x}{3}=\frac{y}{4}\)

đặt K vào \(\frac{x}{3}=\frac{y}{4}\)

ta có

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

\(\frac{y}{4}=K\Rightarrow y=4K\)

\(x.y=48\)

\(3K.4K=48\)

\(12K^2=48\)

\(K^2=48:12=4\)

\(K^2=2^2\Rightarrow K=2\)

*\(\frac{x}{3}=2\Rightarrow x=2.3=6\)

*\(\frac{y}{4}=2\Rightarrow y=2.4=8\)

*\(\frac{z}{7}=2\Rightarrow z=2.7=14\)

vậy \(x=6;y=8;z=14\)

dat \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}=k\) => x=3k,y=4k,z=7k

Thay vvao ta dc: x.y=48

                        3k.4k=48

                        12.\(k^2\)=48

                              k^2=4

                               k=4,-4

TH1: k=a

=> x=3k=>x=12

     y va z lam tuong tu nhe

Con TH2 la -4

k cho m nha

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}\)Và \(x\cdot y=48\)

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}=K\)

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

\(\Rightarrow\frac{y}{4}=K\Rightarrow y=4K\)

\(\Rightarrow\frac{z}{7}=K\Rightarrow z=7K\)

Mà \(x\cdot y=48\)

\(\Rightarrow3K\cdot4k=48\)

\(\Rightarrow12K^2=48\)

\(\Rightarrow K^2=4\)

\(\Rightarrow K=2\)

Khi đó: \(\Rightarrow\frac{x}{3}=2\Rightarrow x=6\)

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

\(\Rightarrow\frac{z}{7}=2\Rightarrow z=14\)

Vậy x=3;y=8 và z=14

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

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

\(\frac{x+1}{3}=\frac{y+2}{4}=\frac{x+3}{5}=\frac{x+y+z+1+2+3}{3+4+5}=\frac{24}{12}=2\)

\(\Rightarrow\)\(\frac{x+1}{3}=2\Rightarrow x=5\)

\(\frac{y+2}{4}=2\Rightarrow y=6\)

\(\frac{z+3}{5}=2\Rightarrow z=7\)

Vậy bạn tự kết luận nha

ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)

\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

ADTCDTSBN

có: \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\frac{x}{8}=2\Rightarrow x=16\)

y/12 = 2 => y = 24

z/15 = 2 => z = 30

KL: x = 16; y=24;z=30

Ta có : 

\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{12}\)

\(\frac{y}{4}=\frac{z}{5}\)\(\Rightarrow\)\(\frac{y}{12}=\frac{z}{15}\)

Suy ra : \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

Do đó : 

\(\frac{x}{8}=2\)\(\Rightarrow\)\(x=2.8=16\)

\(\frac{y}{12}=2\)\(\Rightarrow\)\(y=2.12=24\)

\(\frac{z}{15}=2\)\(\Rightarrow\)\(z=2.15=30\)

Vậy \(x=16\)\(;\)\(y=24\) và \(z=30\)

Chúc bạn học tốt ~ 

Ta có: \(\orbr{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{4}=\frac{z}{5}\end{cases}\Rightarrow\orbr{\begin{cases}\frac{x}{8}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{15}\end{cases}\Rightarrow}\frac{x}{8}=\frac{y}{12}=\frac{z}{15}}\)

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{13}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\) \(\frac{x}{8}=2\Rightarrow x=16\) 

        \(\frac{y}{12}=2\Rightarrow y=24\)

        \(\frac{z}{15}=2\Rightarrow z=30\)

KL:................

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)

\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{14}=\frac{y}{21}\)

\(\frac{y}{7}=\frac{z}{4}\Rightarrow\frac{y}{21}=\frac{z}{12}\)

\(\Leftrightarrow\frac{x}{14}=\frac{y}{21}=\frac{z}{12}=\frac{x+y-z}{14+21-12}=\frac{69}{23}=3\)

\(\Rightarrow x=52;y=63;z=36\)

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{7}=\frac{z}{4}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{14}=\frac{y}{21}\\\frac{y}{21}=\frac{z}{12}\end{cases}\Rightarrow}\frac{x}{14}=\frac{y}{21}=\frac{z}{12}}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{14}=\frac{y}{21}=\frac{z}{12}=\frac{x+y-z}{14+21-12}=\frac{69}{23}=3\)

\(\Rightarrow\hept{\begin{cases}x=3.14=42\\y=3.21=63\\z=3.12=36\end{cases}}\)

Ta có:

\(\frac{x}{2}=\frac{y}{3};\frac{y}{7}=\frac{z}{4}\)

=> \(\frac{x}{14}=\frac{y}{21};\frac{y}{21}=\frac{z}{12}\)

=> \(\frac{x}{14}=\frac{y}{21}=\frac{z}{12}=\frac{x+y-z}{14+21-12}=\frac{69}{23}=3\)

Từ trên ta có:

\(\frac{x}{14}=3=>x=3.14=42\)

\(\frac{y}{21}=3=>y=3.21=63\)

\(\frac{z}{12}=3=>z=3.12=36\)

Vậy x = 42

       y = 63

       z = 36

ỦNG HỘ NHA

 a) Ta có \(\frac{x-1}{2}\)\(=\)\(\frac{y-2}{3}\)\(=\)\(\frac{z-3}{4}\)\(=\)\(\frac{2x-2}{4}\)\(=\)\(\frac{3y-6}{9}\)\(=\)\(\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\)\(=\)\(\frac{\left(2x+3y-z\right)-5}{9}\)\(=\)\(\frac{50-5}{9}\)\(=\)5                                                       Do đó x \(=\)5\(\times\)2\(+\)1\(=\)11                                                                                                                                                           y\(=\)5\(\times\)3\(+\)2\(=\)17                                                                                                                                                            z\(=\)5\(\times\)4\(+\)3\(=\)23