\(x^2+y^4=0\)
\(x^2\ge0;y^4\ge0\)
Dấu "=" xảy ra khi:
\(x^2=0\Rightarrow x=0\)
\(y^4=0\Rightarrow y=0\)
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\)
Dấu"=" xảy ra khi:
\(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
\(\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\)
\(\left(x-11+y\right)^2+\left(x-4-y\right)^2=0\)
\(\left(x-11+y\right)^2\ge0;\left(x-4-y\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left(x-11+y\right)^2=0\Rightarrow x-11+y=0\)
\(\left(x-4-y\right)^2=0\Rightarrow x-4-y=0\)
\(\Rightarrow\left(x-11+y\right)-\left(x-4-y\right)=0\)
\(\Rightarrow x-11+y-x+4+y=0\)
\(\Rightarrow2y-7=0\Rightarrow2y=7\Rightarrow y=\dfrac{7}{2}\)
Thay \(\dfrac{7}{2}\)vào \(2y\) ta có:
\(x-11+y=0\Rightarrow x-11+\dfrac{7}{2}=0\Rightarrow x-\dfrac{15}{2}=0\Rightarrow x=\dfrac{15}{2}\)