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

\(\Rightarrow2x=\frac{5}{12}\)

\(\Rightarrow x=\frac{5}{12}:2\)

\(\Rightarrow x=\frac{5}{24}\)

Có x rồi bạn thế vào => ra được y rồi thế y vòa => được z

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

Áp dụng TC DTSBN ta có :

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

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

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

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

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

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

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

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

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

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

Có : \(\left(x-\frac{1}{x}\right)^2\ge0\)

\(\Leftrightarrow x^2-2+\frac{1}{x^2}\ge0\)

\(\Leftrightarrow x^2+\frac{1}{x^2}\ge2\)

C/m tt \(y^2+\frac{1}{y^2}\ge2\)

Cộng lại ta được \(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}\ge4\)

Dấu "=" khi \(\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)

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

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

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

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

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

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