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

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

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

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

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

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

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

Theo tính chất dãy tỉ số bằng nhau thì:

  \(\frac{x+2}{7}=\frac{y-3}{5}=\frac{z}{3}=\frac{\left(x+2\right)+\left(y-3\right)-x}{7+5-3}=\frac{x+y-z-1}{9}=\frac{-17-1}{9}=-2\)

=> \(\frac{x+2}{7}=-2\Rightarrow x=\left(-2\right).7-2=-16\)

    \(\frac{y-3}{5}=-2\Rightarrow y=\left(-2\right).5+3=-7\)

   \(\frac{z}{3}=-2\Rightarrow z=\left(-2\right).3=-6\)

thưa cô theo em nghĩ thì phải là

\(\frac{x+2}{7}=\frac{y-3}{5}=\frac{z}{3}=\frac{\left(x+2\right)+\left(y-3\right)-z}{7+5-3}\) chứ ạ cô nhầm thì phải ạ

nguyen viet hieu sai roi

a. \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow x=2k;y=3k\)

\(xy=54\Rightarrow2k3k=54\Rightarrow6k^2=54\Rightarrow k^2=9\Rightarrow k\in\left\{3;-3\right\}\)

\(k=3\Rightarrow x=6;y=9\)

\(k=-3\Rightarrow x=-6;y=-9\)

b.\(\frac{x}{5}=\frac{y}{3}=k\Rightarrow x=5k;y=3k\)

\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\Rightarrow25k^2-9k^2=4\)

\(\Rightarrow16k^2=4\Rightarrow k^2=\frac{1}{4}\Rightarrow k\in\left\{\frac{1}{2};-\frac{1}{2}\right\}\)

\(k=\frac{1}{2}\Rightarrow x=\frac{5}{2};y=\frac{3}{2}\)

\(k=-\frac{1}{2}\Rightarrow x=\frac{-5}{2};y=\frac{-3}{2}\)

c.\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{2}.\frac{1}{5}=\frac{y}{3}.\frac{1}{5}\Rightarrow\frac{x}{10}=\frac{y}{15}\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{3}=\frac{z}{7}.\frac{1}{3}\Rightarrow\frac{y}{15}=\frac{z}{21}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)

\(\Rightarrow x=20,y=30,z=42\)

d.\(\frac{x^2}{9}=\frac{y^2}{16}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

\(\Rightarrow x^2=36\Rightarrow x\in\left\{6;-6\right\};y^2=64\Rightarrow y\in\left\{8;-8\right\}\)