Question

a,\(\frac{X}{Y+Z+1}=\frac{Y}{X+Z+3}=\frac{Z}{Y+X-4}=X+Y+Z\)+Z

b,\(\frac{x-18}{x+4}=\frac{x-17}{x+16}\)tìm x

 

Answer 1

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

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

=>\(\hept{\begin{cases}y+z+1=2x\\x+z+3=2y\\x+y-4=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+3=3y\\x+y+z-4=3z\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+3\\3z=\frac{1}{2}-4\end{cases}}}\Rightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{7}{2}\\3z=\frac{-7}{2}\end{cases}}\)

đến đây dễ rồi

b, =>(x-18)(x+16)=(x+4)(x-17)

=>x²+16x-18x-288=x²-17x+4x-68

=>x²-2x-288-x²+13x+68=0

=>11x-220=0

=>11x=220

=>x=20