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\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)(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)\(\left(2\sqrt{x}-3\right)\left(2+\sqrt{x}\right)+6=0\)
\(\Leftrightarrow4\sqrt{x}+2x-6-3\sqrt{x}+6=0\)
\(\Leftrightarrow2x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\2\sqrt{x}-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{4}\end{array}\right.\)
Bài 2:Áp dụng BĐT AM-GM ta có:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\)
\(\frac{1}{y}+\frac{1}{z}\ge2\sqrt{\frac{1}{yz}}\)
\(\frac{1}{x}+\frac{1}{z}\ge2\sqrt{\frac{1}{xz}}\)
CỘng theo vế 3 BĐT trên có:
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
Khi x=y=z
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(..........................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng theo vế ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\)
a) \(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)
b) \(\sqrt{x-26}+\sqrt{y+20}+\sqrt{z+3}=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z-2\sqrt{x-26}-2\sqrt{y+20}-2\sqrt{z+3}=0\)
\(\Leftrightarrow x-26-2\sqrt{x-26}+1+y+20-2\sqrt{y+20}+1+z+3+2\sqrt{z+3}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-26}-1\right)^2+\left(\sqrt{y+20}-1\right)^2+\left(\sqrt{z+3}-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-26}-1=0\\\sqrt{y+20}-1=0\\\sqrt{z+3}-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=27\\y=-19\\z=-2\end{cases}}\)