\(ĐKXĐ:x\ne2009;y\ne2010;z\ne2011;x,y,z\in R\)

\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{y-2010}-\frac{\sqrt{y-2011}}{y-2011}+\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}=\frac{-3}{4}\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}^2}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{y-2010}^2}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{\sqrt{z-2011}^2}+\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^{^2}+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)

• \(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}=0\)

• \(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}=0\)
• \(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{\sqrt{x-2009}}=\frac{1}{2};\frac{1}{\sqrt{y-2010}}=\frac{1}{2};\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\)

\(\Leftrightarrow x=2013;y=2014;z=2015\inĐKXĐ\)

  VẬY       \(x=2013;y=2014;z=2015\)

ko biet E=MC'2

Ta có \(y\ge0\)

\(\Rightarrow P=\left(x^2+2x+1\right)-\left(x\sqrt{y}+\sqrt{y}\right)+y+4\)

\(\Rightarrow P=\left(x+1\right)^2-2.\left(x+1\right).\frac{\sqrt{y}}{2}+\left(\frac{\sqrt{y}}{2}\right)^2+\frac{3y}{4}+4\)

\(\Rightarrow P=\left(\left(x+1\right)-\frac{\sqrt{y}}{2}\right)^2+\frac{3y}{4}+4\)

Vì \(\left(\left(x+1\right)-\frac{\sqrt{y}}{2}\right)^2\ge0;\frac{3y}{4}\ge0\Rightarrow P\ge0+0+4=4\)

vậy minP = 4 khi x = -1 và y = 0

cảm ơn chị

Áp dụng BDT Bunhiacopki, ta có

\(1^2\le\left(x+3y\right)^2\le\left(1^2+3^2\right)\left(X^2+Y^2\right)\)

\(\Rightarrow\)\(X^2+Y^2\)\(\ge\frac{1}{10}\).Dấu bằng xảy ra

                         \(\Leftrightarrow x=3y\)

                          \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{10}\\y=\frac{1}{10}\end{cases}}\)

x^2+x-p=0

=>x^2+x=p

=>x(x+1)=p

Do p là số nguyên tố,mà số nguyên tố chỉ có 2 ước duy nhất là 1 và chính nó

x,x+1 là 2 số tự nhiên liên tiếp=>p=2

2=1.2=(-1).(-2)

với x(x+1)=1.2=>x=1

với x(x+1)=-1.-2=>x=-2

vậy x={1,-2}

Ta co: x²+x-p=0

=>x(x+1)=p

Ma x(x+1) la h cua hai so tu nhien lien tiep

Nen x(x+1)chia het cho 2

Do p la so nguyen to=>p=2

=>x(x+1)=2=1×2=-2×(-1)

=>x =1 hoac x=-2