Question

Cho các số x,y,z thỏa mãn ( Chú ý : A^2+B^2+C^2=0 <=> A=B=C=0)

a, \(\left(2x-y\right)^2+\left(y-2\right)^2+\sqrt{x+y+z}=0\)

b, \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

Answer 1

a) ta thấy : \(\left(2x-y\right)^2>=0;\left(y-2\right)^2>=0;\sqrt{x+y+z}>=0\)

mà \(\left(2x-y\right)^2+\left(y-2\right)^2+\sqrt{x+y+z}=0\)

=> \(\left(2x-y\right)^2=0\)

        \(\left(y-2\right)^2=0\)

      \(\sqrt{x+y+z}=0\)

=> y=2;x=1;z=-3

Answer 2

b) ta có :

\(x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\)

=> \(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

<=> \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

=>  \(\left(\sqrt{x-2}-1\right)^2=0;\left(\sqrt{y-3}-2\right)^2=0;\left(\sqrt{z-5}-3\right)^2=0\)

=> x=3;y=7;z=14