Question

tìm x,y,z biết

a) x+y+z+12=4\(\sqrt{x}+6\sqrt{y-1}\)

b)x+y+z+8=2\(\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)

c)\(\sqrt{x-2001}+\sqrt{x-2002}-\sqrt{x-2003}=\dfrac{1}{2}\left(x+y+z\right)-3015\)

Answer 1

hình như...

b) \(x+y+z+8=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)

\(\Leftrightarrow x-3+y-3+z-3+17=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)

\(\Leftrightarrow\left(x-3-2\sqrt{x-3}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)+3=0\)

\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-3}-3\right)^2+3=0\) (vô nghiệm, VT >/3)

Kl: ptvn

Answer 2

c) là y - 2002 , z-2003 chứ 0 phải x đúng 0? (đoán thôi)