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Phương trình có vô số nghiệm

Nếu thay \(\sqrt{y-2008}\) bằng \(\sqrt{y+2008}\) thì phương trình có bộ nghiệm duy nhất: \(\left(x;y;z\right)=\left(2010;-2007;3\right)\)

\(x-2008=X;y-2009=Y;z-2010=Z\)

\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)

\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)

\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)

\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)

\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)

\(\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{\sqrt{z-2011}-1}{z-2011}=\dfrac{3}{4}\)\(\left(\left\{{}\begin{matrix}x>2009\\y>2010\\z>2011\end{matrix}\right.\right)\)

\(\Leftrightarrow\dfrac{1}{4}-\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{1}{4}-\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{1}{4}-\dfrac{\sqrt{z-2011}-1}{z-2011}=0\)

\(\Leftrightarrow\dfrac{x-2009-4\sqrt{x-2009}+4}{x-2009}+\dfrac{y-2010-4\sqrt{y-2010}+4}{y-2010}+\dfrac{z-2011-4\sqrt{z-2011}+4}{z-2011}=0\)

Nhận xét: \(\left\{{}\begin{matrix}\dfrac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}\ge0\\\dfrac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}\ge0\\\dfrac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(2013;2014;2015\right)\)

\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)(đk;x>0)

\(\Leftrightarrow x^2+2\sqrt{x}\cdot\sqrt{x^2+1}=8x-1\)

\(\Leftrightarrow\left(x^2+1\right)+2\sqrt{x}\cdot\sqrt{x^2+1}+x=9x\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+\sqrt{x}\right)^2-9x=0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+\sqrt{x}+3\sqrt{x}\right)\left(\sqrt{x^2+1}+\sqrt{x}-3\sqrt{x}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+4\sqrt{x}\right)\left(\sqrt{x^2+1}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\sqrt{x^2+1}-2\sqrt{x}=0\)(vì \(\sqrt{x^2+1}+4\sqrt{x}>0\))

\(\Leftrightarrow x^2-4x+1=0\)

\(\Leftrightarrow\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2-\sqrt{3}\\x=2+\sqrt{3}\end{cases}}\)(thõa mãn điều kiện)

\(\sqrt{x-2009}-\sqrt{y-2008}-\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)(đk:x>2009,y>2008,z>2)

\(\Leftrightarrow\left(\sqrt{x-2009}-1\right)^2+\left(\sqrt{x-2008}+1\right)^2+\left(\sqrt{z-2}+1\right)^2+4014=0\)(không thõa mãn)

Lý do có kết quả trên là vì chuyển 1\2 qua vế trái và tách theo hằng đẳng thức

Bài tiếp theo cũng làm tương tự

a) \(\sqrt{25-x^2}-\sqrt{10-x^2}=3\) (*)

Đk: \(-\sqrt{10}\le x\le\sqrt{10}\)

(*) \(\Leftrightarrow\sqrt{25-x^2}=3+\sqrt{10-x^2}\Leftrightarrow25-x^2=19-x^2+6\sqrt{10-x^2}\)

\(\Leftrightarrow6\sqrt{10-x^2}=6\Leftrightarrow\sqrt{10-x^2}=1\Leftrightarrow\left[{}\begin{matrix}x=-3\left(N\right)\\x=3\left(N\right)\end{matrix}\right.\)

Kl: x = +- 3

b) \(\sqrt{x^2-x-6}+x^2-x-18=0\) (*)

đk: \(\left[{}\begin{matrix}x\le-2\\x\ge3\end{matrix}\right.\)

(*) \(\Leftrightarrow x^2-x-6+\sqrt{x^2-x-6}-12=0\)

Đặt \(t=\sqrt{x^2-x-6}\Rightarrow t^2=x^2-x-6\) (t >/ 0)

phương trình (*) trở thành : \(t^2+t-12=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(N\right)\\t=-4\left(L\right)\end{matrix}\right.\)

Với t=3. ta có: \(\sqrt{x^2-x-6}=3\Leftrightarrow x^2-x-15=0\Leftrightarrow x=\dfrac{1\pm\sqrt{61}}{2}\left(N\right)\)

Kl: \(x=\dfrac{1\pm\sqrt{61}}{2}\)

c) \(\sqrt{x-2009}+\sqrt{y+2008}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\) (*)

Đk: \(\left\{{}\begin{matrix}x\ge2009\\y\ge-2008\\z\ge2\end{matrix}\right.\)

(*) \(\Leftrightarrow2\sqrt{x-2009}+2\sqrt{y+2008}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow\left(x-2009-2\sqrt{x-2009}+1\right)+\left(y+2008-2\sqrt{y+2008}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2009}-1\right)^2+\left(\sqrt{y+2008}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}=1\\\sqrt{y+2008}=1\\\sqrt{z-2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2010\left(N\right)\\y=-2007\left(N\right)\\z=3\left(N\right)\end{matrix}\right.\)

Kl: x= 2010, y= -2007, z=3