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9x+2=y^2+y<=>9x+2=y(y+1) 

VT chia 3 dư 2 nên VP cũng chia 3 dư 2 

=>y=3k+1 và y+1=3k+2 ( k nguyên)

khi đó 9x+2=(3k+1)(3k+2) 

<=>9x=9k^2+9k <=>x=k^2+k<=>x=k(k+1) 

Vậy pt có nghiệm nguyên (x;y) là (k(k+1));(3k+1) 

a) x - \(\sqrt{x^2-2x+1}\)+ 2

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

= x - Ix-1I + 2 ( * )

= x - x - 1 + 2 (vì x >1)

= -1

b) Thế x = \(\frac{1}{2}\)vào ( * ), ta dc: \(\frac{1}{2}\)- I\(\frac{1}{2}\)- 1 I + 2 = \(\frac{1}{2}\)-  ( 1 - \(\frac{1}{2}\)) + 2 = 2

Ta có : P = \(1-3x+\frac{3}{2-x}=6-3x+\frac{3}{2-x}\) \(-5\) \(=3\left(2-x\right)+\frac{3}{2-x}-5\)

Áp dụng BĐT : AM-GM ta được :

\(3\left(2-x\right)+\frac{3}{2-x}\ge2\sqrt{\frac{3\left(2-x\right)3}{\left(2-x\right)}}=\)\(2\sqrt{9}=2.3=6\)

Vì x<2 => dấu "=" xảy ra khi : x=1

=> P \(\ge6-5=1\)

Vậy Min P = 1 khi x=1

\(\frac{x^2-5}{3x}-\frac{2}{3}=\frac{x^2-2x-5}{3x}=\frac{\left(x-1+\sqrt{6}\right)\left(x-1-\sqrt{6}\right)}{3x}\)

Đề bài kiểu gì vậy?