Question

Giải các phương trình sau:

a) \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2003}=\frac{x-12}{2003}+2\)

b) \(\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)

Answer 1

a. \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2003}=\frac{x-12}{2003}+2\)

\(\rightarrow\frac{x}{2000}-\frac{15}{2000}+\frac{x}{2001}-\frac{14}{2001}+\frac{x}{2003}-\frac{13}{2003}=\frac{x}{2003}-\frac{12}{2003}+2\)

\(\rightarrow x.\left(\frac{1}{2000}+\frac{1}{2001}\right)=\frac{15}{2000}+\frac{14}{2001}+\frac{13}{2003}-\frac{12}{2003}+2\)

\(\rightarrow x=2015,5\)

b. \(\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)

\(\rightarrow\left\{{}\begin{matrix}x^2-6x+11=\left(x-3\right)^2+2\ge2\\y^2+2y+4=\left(y+1\right)^2+3\ge3\\2+4z-z^2=-\left(z-2\right)^2+6\le6\end{matrix}\right.\)

\(\rightarrow\left(x^2-6x+11\right)\left(y^2+2y+4\right)\ge6\)

\(\rightarrow\left(x^2-6x+11\right)\left(y^2+2y+4\right)=2+4z-z^2\)

\(\rightarrow\left\{{}\begin{matrix}x=3\\y=-1\\z=2\end{matrix}\right.\)