căng nha

\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)

\(=\left[\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}< 1\)

\(\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{\frac{1}{x+y+x}}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)=1\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Leftrightarrow\left(x+y\right)\left[z\left(x+y+z\right)+xy\right]=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

B=\(\left(x+y\right)\left(y+z\right)\left(z+x\right).M=0\)

M ở đâu ra thế bạn

\(x^2-2xy+x-2y\le0\Leftrightarrow x\left(x-2y\right)+\left(x-2y\right)\le0\Leftrightarrow\left(x+1\right)\left(x-2y\right)\le0\)

Vì \(x\ge0\Rightarrow x+1\ge0\Rightarrow x-2y\le0\Rightarrow x\le2y\)

\(A=x^2-5y^2+3x\le\left(2y\right)^2-5y^2+3.2y=-y^2+6y=9-\left(y-3\right)^2\le9\)

=>\(A\le9\)

Dấu "=" xảy ra khi x=6;y=3

BĐT đã cho tương đương với :

\(\left(\frac{a^2}{b}+\frac{b^2}{a}\right)^2\ge2\left(a^2+b^2\right)\)

\(\Leftrightarrow\frac{\left(a^3+b^3\right)^2}{\left(ab\right)^2}\ge2\left(a^2+b^2\right)\Leftrightarrow\frac{\left(a^3+b^3\right)^2-2\left(a^2+b^2\right)\left(ab\right)^2}{\left(ab\right)^2}\ge0\)

\(\Leftrightarrow\frac{a^6+b^6+2a^3b^3-2a^4b^2-2a^2b^4}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{a^4\left(a^2-b^2\right)+b^4\left(b^2-a^2\right)+2a^3b^3-a^4b^2-a^2b^4}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{\left(a^2-b^2\right)\left(a^4-b^4\right)+a^2b^2\left(2ab-a^2-b^2\right)}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(a+b\right)^2\left(a^2+b^2\right)-a^2b^2\left(a-b\right)^2}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left[\left(a+b\right)^2\left(a^2+b^2\right)-a^2b^2\right]}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(a^4+b^4+a^2b^2+2a^3b+2ab^3\right)}{a^2b^2}\ge0\)

Dấu "=" xảy ra khi a = b  khác 0

vì nó là số chính phương đặt 

\(n^2+n+503=a^2\Leftrightarrow4n^2+4n+1+2011=4a^2\)

<=>\(\left(2n+1\right)^2+2011=4a^2\)

<=> \(4a^2-\left(2n+1\right)^2=2011\)

<=> \(\left(2a-2n-1\right)\left(2a+2n+1\right)=2011\)

tìm ước rồi lập bảng chắc 2011 cs ít ước thôi

Đặt \(n^2+n+503=k^2\left(k\in N\right)\Leftrightarrow4n^2+4n+2012=4k^2\Leftrightarrow\left(2n+1\right)^2\)+2011=(2k)²

\(\Leftrightarrow\left(2k\right)^2-\left(2n+1\right)^2=2011\Leftrightarrow\left(2k-2n-1\right)\left(2k+2n+1\right)=2011\)

<=>2k-2n-1=-1 và 2k+2n+1=-2011 hoặc

      2k-2n-1=1 và 2k+2n+1=2011

giải 2 cái hệ đó ra thì tìm được n