Tìm \(x;y\in N\)tmãn : \(\sqrt{x}+\sqrt{y}=\sqrt{2012}\)
2, Rút gọn bt
\(P=\dfrac{x}{x-\sqrt{x}}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
b, gpt : \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)
3, cho x>1 ; y>0 , cm
\(\dfrac{1}{\left(x+1\right)^3}+\left(\dfrac{x-1}{y}\right)^3+\dfrac{1}{y^3}\ge3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)
Unruly Kid
Đặt VT là T
Áp dụng AM-GM cho 3 số dương, ta có:
\(\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)
\(T\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)(đpcm)
\(P=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{.....}+\dfrac{x+2}{....}\)
\(=\dfrac{\sqrt{x^3}+2x+2\sqrt{x}-2+x+2}{.....}=\dfrac{\sqrt{x^3}+3x+2\sqrt{x}}{....}\)
\(=\dfrac{\sqrt{x}\left(x+3\sqrt{x}+2\right)}{....}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{....}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P/S: Chú ý điều kiện khi rút gọn, tự tìm.
2)
P = \(\dfrac{x}{x-\sqrt{x}}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\) với \(x>0;x\ne1\)
\(\Rightarrow P=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
= \(\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)= \(\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-1\right)+\left(x+2\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
= \(\dfrac{x\sqrt{x}+2x+2\sqrt{x}-2+x+2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{x\sqrt{x}+3x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) ĐK: \(x\ge0\)
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)
\(\Leftrightarrow x^2+2\sqrt{x^3}-2\sqrt{x}-x\sqrt{x}-2\sqrt{x^3}-4x+4+2x=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x^3}+2x-2-x\right)-2\left(\sqrt{x^3}+2x-2-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x^3}+x-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+2\right)=0\)
Dễ thấy \(x+2\sqrt{x}+2>0\)
Suy ra: \(\left[{}\begin{matrix}\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\\\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\end{matrix}\right.\)(TMĐK)
Vậy: \(S=\left\{1;4\right\}\)