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1) Ta có: \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
Áp dụng liên tiếp bđt Cauchy-Schwarz và AM-GM
\(\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}=\dfrac{x^2}{x+y^2x}+\dfrac{y^2}{y+x^2y}\)
\(\ge\dfrac{\left(x+y\right)^2}{x+y+y^2x+x^2y}=\dfrac{4}{x+y+xy\left(x+y\right)}\)
\(=\dfrac{4}{2+2xy}\ge\dfrac{4}{2+\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{4}=1\)
\("="\Leftrightarrow x=y=1\)
Q= \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\) (x≥0, x≠4)
Q= \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}+\dfrac{5\sqrt{x}-2}{x-4}\)
Q= \(\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}\)
Q= \(\dfrac{x+2\sqrt{x}}{x-4}\)
Q= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)
\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)
\(A=\frac{x^2+2004x+2003}{x^2-x}\)
\(M=\left(\frac{2x}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(M=\left(\frac{2x}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right):\frac{x+1+\sqrt{x}}{x+1}\)
\(M=\frac{2x}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\frac{x+\sqrt{x}+1}{x+1}\)
\(M=\frac{2x}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\frac{x+1}{x+\sqrt{x}+1}\)
\(M=\frac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(M=\frac{2x}{x^3-1}\)
vậy \(M=\frac{2x}{x^3-1}\)