Question

Tìm các số \(x,y,z\)biết rằng:

a) \(\frac{x}{5}=\frac{y}{3}\)và \(x^2-y^2=4\)

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

Answer 1

- Vì \(\frac{x}{5}=\frac{y}{3}\)=) \(3x=5y\)=) \(x=\frac{5y}{3}\)
=) \(x^2-y^2=4\)=) \(\left(\frac{5y}{3}\right)^2-y^2=4\)
=) \(\frac{25y^2}{9}-y^2=4\)=) \(\frac{25y^2}{9}-\frac{9y^2}{9}=\frac{36}{9}\)
=) \(25y^2-9y^2=36\)=) \(16y^2=36\)=) \(y^2=\frac{36}{16}=\frac{9}{4}\frac{3^2}{2^2}\)=) \(y=\frac{3}{2}\)
=) \(x=\frac{5.\frac{3}{2}}{3}=\frac{\frac{15}{2}}{3}=\frac{5}{2}\)

Answer 2

a) Đặt x/5 = y/3 = k => x = 5k ; y = 3k

Ta có: x² - y² = 4

=> (5k)² - (3k)² = 4

=> 25k² - 9k² = 4

=> 16k² = 4

=> k² = 1/4

=> k = ±1/2

Với k = 1/2 thì x = 5/2, y = 3/2

Với k = -1/2 thì x = -5/2, y = -3/2

b) Theo tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+z+x+1+x+y-2}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

=> x + y + z = 1/2 ; x/y+z+1 = 1/2 ; y/z+x+1 = 1/2 ; z/x+y-2 = 1/2

=> \(\hept{\begin{cases}y+z+1=2x\\z+x+1=2y\\x+y-2=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+1=3y\\\frac{1}{2}-2=3z\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{cases}}\)