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\(DK:x\ge1\)
\(A=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}+2019\)
\(=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|+2019\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|+2019\ge|\sqrt{x-1}+1+1-\sqrt{x-1}|+2019=2021\)
Dau '=' xay ra khi \(\left(\sqrt{x-1}+1\right)\left(1-\sqrt{x-1}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x-1}+1\ge0\\1-\sqrt{x-1}\ge0\end{cases}\Leftrightarrow x=2\left(n\right)}\)
TH2:
\(\hept{\begin{cases}\sqrt{x-1}+1\le0\\1-\sqrt{x-1}\le0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}\le-1\\\sqrt{x-1}\ge1\end{cases}\left(l\right)}}\)
Vay \(A_{min}=2021\)khi \(x=2\)
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{2\sqrt{x}+x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)
\(A=\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{14\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{14\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{2x+5\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-12\sqrt{x}-13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-13\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-13}{\sqrt{x}+3}\)
\(A=\frac{\sqrt{x}+3-16}{\sqrt{x}+3}=1-\frac{16}{\sqrt{x}+3}\ge1-\frac{16}{3}=-\frac{13}{3}\)
\(A_{min}=-\frac{13}{3}\) khi \(x=0\)
+) \(B=6\sqrt{x-2}+6\sqrt{5-x}\Leftrightarrow B^2=\left(6\sqrt{x-2}+6\sqrt{5-x}\right)^2\)
\(=36\left(x-2\right)+36\left(5-x\right)+72\sqrt{\left(x-2\right)\left(5-x\right)}\ge108\Rightarrow B\ge6\sqrt{3}\)
+) \(A=B+2\sqrt{5-x}\ge6\sqrt{3}\)
Vậy \(A_{min}=6\sqrt{3}\)khi x=5
+) Đặt \(a=\sqrt{x-2};b=\sqrt{5-x}\)
+) Ta có: \(a^2+b^2=3\)
+) \(\left(a^2+b^2\right)\left(6^2+8^2\right)\ge\left(6a+8b\right)^2\Leftrightarrow\left(6a+8b\right)^2\le300\Rightarrow6a+8b\le10\sqrt{3}\)
Dấu = xảy ra khi \(\frac{a}{6}=\frac{b}{8}\Leftrightarrow\frac{\sqrt{x-2}}{6}=\frac{\sqrt{5-x}}{8}\Leftrightarrow\frac{x-2}{36}=\frac{5-x}{64}\Leftrightarrow64x-128=180-36x\Leftrightarrow308=100x\)
\(\Leftrightarrow x=3.08\)
Vậy \(A_{max}=10\sqrt{3}\)khi x=3.08