15+25+35+45+55+65+75+85=[15+85]+[25+75]+[35+65]+[45+55]

                                               =100+100+100+100=400

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xét vs x=0,1,2 ko tm
vs x=3 =>y=3
ta di c/m đây là no duy nhất.
thật vậy vs x>3

\(x^2+7x+10=\left(x^2+2x\right)+\left(5x+10\right)=x\left(x+2\right)+5\left(x+2\right)=\left(x+2\right)\left(x+5\right)\)

thay x = 97 vào đc:

(97 + 2) (97 + 5) = 99 . 102 = 10098

??? ko thấy cái j là nhanh ~0~...

Với điều kiện  \(0\)\(\le x\le1\)ta có

P= \(\sqrt{\frac{\sqrt{x\left(x\sqrt{x-1}\right)}}{x+\sqrt{x}+1}-\sqrt{\frac{\sqrt{x\left(x\sqrt{x+1}\right)}}{x-\sqrt{x}+1}+x+1}}\)

sử dụng hằng đẳng thức bậc 3  : \(x^3\)- \(y^3\)và \(x^3\)+  \(y^3\)

ta có P = \(\sqrt{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1}\)= \(\sqrt{x-2\sqrt{x}+1}=\)\(\sqrt{\left(\sqrt{x}-1\right)^2}\)=\(\left|\sqrt{x}-1\right|\)

P= \(1-\sqrt{x}\)

\(2< \sqrt{6}< 3.\)

\(2< \sqrt{6+2}< \sqrt{6+\sqrt{6}}< \sqrt{6+3}=3\)

\(2< \sqrt{6+2}< \sqrt{6+\sqrt{6+\sqrt{6}}}< \sqrt{6+3}=3\)

...

\(2< \sqrt{6+2}< \sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}< \sqrt{6+3}=3\)

Vậy phần nguyên của \(A=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}\)là 2

e, \(A=\frac{2010^{2010}+1}{2010^{2011}}=>2010A=\frac{2010.\left(2010^{2010}+1\right)}{2010^{2011}}=\frac{2010^{2011}+2010}{2010^{2011}}=1+\frac{2010}{2010^{2011}}\)

   \(B=\frac{2010^{2011}+1}{2010^{2012}}=>2010B=\frac{2010^{2012}+2010}{2010^{2012}}-1+\frac{2010}{2010^{2012}}\)

Vì \(2010^{2011}< 2010^{2012}=>\frac{2010}{2010^{2011}}>\frac{2010}{2010^{2012}}\)

=>2010A > 2010B

=> A > B

k cho mk nhé mấy câu trước dễ nên ko làm

 \(\sqrt{1-xy}=\frac{\sqrt{1-xy}.x^2y^2}{x^2y^2}\)\(=\frac{\sqrt{x^4y^4-x^5y^5}}{x^2y^2}\)
có: \(x^5+y^5=2x^2y^2\Rightarrow x^2y^2=\frac{x^5+y^5}{2}\)
\(\frac{\sqrt{x^4y^4-x^5y^5}}{x^2y^2}=\frac{\sqrt{\left(\frac{x^5+y^5}{2}\right)^2-x^5y^5}}{x^2y^2}=\frac{\sqrt{\left(x^5-y^5\right)^2}}{2x^2y^2}=\frac{\left|x^5-y^5\right|}{2x^2y^2}\)
Do x, y hữu tỉ nên \(\frac{\left|x^5-y^5\right|}{2x^2y^2}\)hữu tỉ (đpcm)

c) \(\left(2x^2-5x+3\right)^2=\left(x^2+x-2\right)^2\)

\(\Rightarrow2x^2-5x+3=x^2+x-2\)

\(2x^2-x^2-5x-x=2-3\)

\(x^2-6x=-1\)

\(x\left(x-6\right)=-1\)

ko bt nữa...

b) \(x^{16}+x^8+1=0\)

\(x^8\left(x^8+1\right)+1=0\)

\(x^8\left(x^8+1\right)=-1\)

Ta có: x^8 >/  0 

x^8 + 1  >/  1

=> x^8(x^8 +1)   >/   0

Vậy x thuộc rỗng.