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

dạ mơn nha

A=|x²+y²+5+2x-4y|-|-(x+y-1)²|+2xy

<=>A=||(x²+2x+1)+(y²-4y+4)| - (x+y-1)² + 2xy

= |(x+1)²+(y-2)²| - (x+y-1)² + 2xy

= (x+1)²+(y-2)²-(x+y-1)²+2xy

Đặt x+1=a và y-2=b

=> A = a² + b² - (a+b)² + 2(a-1).(b+2)

= a² + b² - a² - 2ab - b² - 2ab + 4a - 2b - 2

= 4a - 2b - 2

= 4(x + 1)-2(y-2)-2

= 4x+4-2y-4-2

= 4x-2y-2

Thay x = 2²⁰¹⁹ và y = 16⁵⁰³ = 2²⁰¹² vào A, ta có:

A = 4.2²⁰¹⁹ - 2.2²⁰¹² - 2

= 2²⁰²¹ - 2²⁰¹³ - 2

tích đúng mình làm cho

là sao ạk
giải giùm mình với ạk

1/Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay x-y=7 vào A ta đc:

\(A=7^2+2.7+37=49+14+37=100\)

2/Ta có :\(B=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

Thay x+2y=5 vào B ta đc

\(B=5^2-2.5+10=25-10+10=25\)

\(x^2+4y^2-5x+10y-4xy+20\)

\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)

\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)

\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được : 

\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)

\(B=x^2-2xy-2x+2y+y^2\)

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

\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được : 

\(=1-2=-1\)

a)  \(A=x^2+2xy+y^2-4x-4y+1\)

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

\(=3^2-4.3+1=-2\)

b)  \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+37=100\)

c)  \(C=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2.5+10=25\)

a) \(A=x^2+2xy+y^2-4x-4v+1\)

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

\(=3^2-4.3+1=-2\)

\(\frac{2011^3+11^3}{2011^3+2000^3}=\frac{\left(2011+11\right)\left(2011^2+11^2-11.2011\right)}{\left(2011+200\right)\left(2011^2+2000^2-2000.2011\right)}\)

Cần chứng minh \(2011^2+11^2-2011.11=2011^2+2000^2-2000.2011\)

Điều này không khó.

\(B=1-\frac{2}{x}+\frac{2011}{x^2}=2011t^2-2t+1\text{ (với }t=\frac{1}{x}\text{)}\)

->Gộp hằng đẳng thức....

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

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

\(=4x-2y+4\)

thay số.Lưu ý: \(y=16^{503}=\left(2^4\right)^{503}=2^{2012}\)