Ta có: \(\sqrt{\left(x-\sqrt{2}\right)^2}\ge0;\sqrt{\left(y+\sqrt{2}\right)^2}\ge0;\left|x+y+z\right|\ge0\)
Mà theo đề: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
=> \(\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{\left(y+\sqrt{2}\right)^2}=\left|x+y+z\right|=0\)
=> \(x-\sqrt{2}=y+\sqrt{2}=x+y+z=0\)
=> \(x=\sqrt{2};y=-\sqrt{2};z=0\).
tim x biet
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y-\sqrt{2}\right)^2}\)+Ix+y+zI=0
tìm các số x,y,z thỏa mãn đẳng thức :
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+Ix+y+zI=0\)
tim x biet
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
bai 1 :
biet do thi ham so y=\(\left(\sqrt{a}-3\right)x\) di qua \(N\left(\sqrt{5};\sqrt{5}\right)\)( a la hang so)
vay a=?
Bai 2:
Cho \(Q\left(x\right)=ax^2+bx+c\)thoa man :
\(Q\left(0\right)=1;Q\left(1\right)=6;Q\left(2\right)=5\)Tinh b
bai 3 :tinh gia tri bieu thuc
\(A=3x+2y+z\)biet \(\left(x-3y^2\right)+\left(y-1\right)^2+\left(x+z\right)^2=0\)
tim x,y,z biet \(\sqrt{\left(x-\sqrt{5}\right)^2}+\sqrt{\left(y+\sqrt{3}\right)^2}+\left|x-y-z\right|\)
\(\sqrt{\left(x-2\right)^2}+\sqrt{\left(y+2\right)^2}+\left|x+y+z\right|=0\)
tim x,y,z
Tìm x;y;z biết \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
Tìm các số x,y,z thỏa mãn đẳng thức:\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)| = 0
Tìm x,y,z
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(x-\sqrt{2}\right)^2}+\left(x+y+z=0\right)\)
