Question

tính giá trị biểu thức:

\(\left|x^2+y^2+5+2x-4y\right|-\left|-\left(x+y-1\right)^2\right|+2xy.\)

với \(x=2^{2011}\)và \(y=16^{503}\)

Answer 1

Ta có:

         \(x^2+y^2+5+2x-4y\)

\(=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)\)

\(=\left(x+1\right)^2+\left(y-2\right)^2\)\(>0\)

\(\Rightarrow\)\(\left|x^2+y^2+5+2x-4y\right|=\left(x+1\right)^2+\left(y-2\right)^2\)

         \(-\left(x+y-1\right)^2\)\(< 0\)

\(\Rightarrow\)\(\left|-\left(x+y-1\right)^2\right|=\left(x+y-1\right)^2\)

     \(\left|x^2+y^2+5+2x-4y\right|-\left|-\left(x+y-1\right)^2\right|+2xy\)

\(=\left(x+1\right)^2+\left(y-2\right)^2-\left(x+y-1\right)^2+2xy\)  

\(=4x-2y+4\)   (rút gọn nha)

\(=4.2^{2011}-2.16^{503}+4\)

\(=2^{2013}-2^{2013}+4=4\)

P/s:  bn tham khảo nhé, mk ko biết đúng or sai, lm bừa

Answer 2

dạ mơn nha