Question

[nguyễn Trung kiên]

Tìm x; y để (x-1/2+y)^2+(x+5/6)^2=0

Answer 1

[Trần Tuyết Như]

\(\left(x-\frac{1}{2}+y\right)^2+\left(x+\frac{5}{6}\right)^2=0\)

mà: \(\left(x-\frac{1}{2}+y\right)^2\ge0;\left(x+\frac{5}{6}\right)^2\ge0\)

=> để: \(\left(x-\frac{1}{2}+y\right)^2+\left(x+\frac{5}{6}\right)^2=0\)

thì: \(\left(x-\frac{1}{2}+y\right)^2=\left(x+\frac{5}{6}\right)^2=0\)

• \(\left(x+\frac{5}{6}\right)^2=0\Rightarrow x+\frac{5}{6}=0\Rightarrow x=0-\frac{5}{6}=-\frac{5}{6}\)
• \(\left(x-\frac{1}{2}+y\right)^2=0\Rightarrow x-\frac{1}{2}+y=0\Rightarrow\frac{-5}{6}-\frac{1}{2}+y=0\Rightarrow y=0+\frac{5}{6}+\frac{1}{2}=\frac{4}{3}\)

=>   x = -5/6; y = 4/3