Question

Tìm x, y biết:

a) \(\frac{x}{2}=\frac{y}{5}\) và x.y =10

b) 7x = 3y và x - y = 16

c) \(\frac{x+1}{x-1}=\frac{x-2}{x+3}\)

Answer 1

a) Đặt \(\frac{x}{2}=\frac{y}{5}=k\)

\(\Rightarrow x=2k\)

\(\Rightarrow y=5k\)

\(\Rightarrow xy=2k.5k=10k^2\)

\(\Rightarrow10k^2=10\)

\(\Rightarrow k^2=\frac{10}{10}=1\Rightarrow\left[\begin{array}{nghiempt}k=1\\k=-1\end{array}\right.\)

Với \(k=1\)

\(\Rightarrow x=2k=2.1=2\)

\(\Rightarrow y=5k\Rightarrow y=5.1=5\)

Với \(k=-1\)

\(\Rightarrow x=2k=-1.2=-2\)

\(\Rightarrow y=5k=-1.5=-5\)

 

Answer 2

b) \(7x=3y\Rightarrow\frac{7x}{21}=\frac{3y}{21}\Rightarrow\frac{x}{7}=\frac{y}{3}\)

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

\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{16}{4}=4\)

\(x=4.7=28\)\(y=4.3=12\)

Vậy: \(x=28,y=12\)

Answer 3

c) \(\frac{x+1}{x-1}=\frac{x-2}{x+3}\)

\(\Rightarrow\left(x+1\right)\left(x+3\right)=\left(x-1\right)\left(x-2\right)\)

\(\Rightarrow x^2+3x+x+3=x^2-2x-x+2\)

\(\Rightarrow x^2+4x+3=x^2-3x+2\)

\(\Rightarrow x^2-x^2+3=-3x-4x+2\)

\(\Rightarrow7x=-1\)

\(\Rightarrow x=-\frac{1}{7}\)