Question

\(\left(\frac{1}{2}x-5\right)^2+\left(y^2-\frac{1}{4}\right)^4=0\)

Answer 1

\(\left(\frac{1}{2}x-5\right)^2+\left(y^2-\frac{1}{4}\right)^4=0\)

\(\frac{1}{2}x-5=y^2-\frac{1}{4}=0\)

\(\frac{1}{2}x=5;x=10\)

\(y^2-\frac{1}{4}=0;y^2=\frac{1}{4};y=\frac{1}{2}\)

Answer 2

 \(\left(\frac{1}{2}x-5\right)^2+\left(y^2-\frac{1}{4}\right)^4=0\)

Ta có: 

 \(\left(\frac{1}{2}x-5\right)^2\ge0;\left(y^2-\frac{1}{4}\right)^4\ge0\)

=> để  \(\left(\frac{1}{2}x-5\right)^2+\left(y^2-\frac{1}{4}\right)^4=0\)  thì:

       \(\frac{1}{2}x-5=y^2-\frac{1}{4}=0\)

\(\frac{1}{2}x-5=0\Rightarrow\)\(\frac{1}{2}x=5\Rightarrow x=5:\frac{1}{2}=10\)

\(y^2-\frac{1}{4}=0\Rightarrow y^2=\frac{1}{4}\Rightarrow y=\frac{1}{2}\)

Answer 3

y=1/2 chắc luôn cho tớ nhé