Question

Tìm x, y, z:
\(a,\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\\ b,\dfrac{12x-15y}{7}=\dfrac{20z-15x}{9}=\dfrac{15y-20z}{11}\\ va\\ x+y+z=48\)

Answer 1

a: =>x^2+2x-3=x^2-4

=>2x=-1

=>x=-1/2

b: \(\dfrac{12x-15y}{7}=\dfrac{20z-15x}{9}=\dfrac{15y-20z}{11}\)

\(=\dfrac{12x-15y+20z-15x+15y-20z}{7+9+11}=\dfrac{-3x}{27}=\dfrac{-x}{9}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-15y}{7}=\dfrac{-x}{9}\\\dfrac{20z-15x}{9}=\dfrac{-x}{9}\\\dfrac{15y-20z}{11}=\dfrac{-x}{9}\\x+y+z=48\end{matrix}\right.\)

\(\Leftrightarrow\begin{matrix}-115x+135y=0\\20z-14x=0\\135y-180z+11x=0\\x+y+z=48\end{matrix}\)

=>\(\left(x,y,z\right)\in\varnothing\)