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a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)
\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)
\(Q=\dfrac{1}{x-2\sqrt{x}+3}\)
Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)
\(a,P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{4-x}\right):\dfrac{x+5\sqrt{x}+6}{x-4}\left(dk:x\ge0,x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{4x}{x-4}\right).\dfrac{x-4}{x+2\sqrt{x}+3\sqrt{x}+6}\)
\(=\dfrac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2+4x}{x-4}.\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x+8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+3}\)
\(b,x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{4}}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\\ =\sqrt{5}+2-\sqrt{5}+2\\ =4\)
Khi \(x=4\Rightarrow P=\dfrac{4\sqrt{4}}{\sqrt{4}+3}=\dfrac{4.2}{2+3}=\dfrac{8}{5}\)
\(c,P=2\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+3}=2\Leftrightarrow\dfrac{4\sqrt{x}-2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=0\Leftrightarrow2\sqrt{x}-6=0\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
a)Đặt \(A=\sqrt{x-2}+\sqrt{4-x}\)
Đk:\(2\le x\le4\)
\(A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}\)
\(=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\) (dùng BĐT Cauchy)
\(\le2+\left(x-2\right)+\left(4-x\right)\)
\(=2+2=4\)
\(\Rightarrow A^2\le4\Leftrightarrow A\le2\)
Dấu = khi \(\sqrt{x-2}=\sqrt{4-x}\Leftrightarrow x=3\)
Vậy MaxA=2 khi x=3
b)Đặt \(B=\sqrt{6-x}+\sqrt{x+2}\)
Đk:\(-2\le x\le6\)
\(B^2=6-x+x+2+2\sqrt{\left(6-x\right)\left(x+2\right)}\)
\(=8+2\sqrt{\left(6-x\right)\left(x+2\right)}\) (Bđt Cauchy)
\(\le8+\left(6-x\right)+\left(x+2\right)\)
\(=8+8=16\)
\(\Rightarrow B^2\le16\Leftrightarrow B\le4\)
Dấu = khi \(\sqrt{6-x}=\sqrt{x+2}\Leftrightarrow x=2\)
Vậy MaxB=4 khi x=2
c)Đặt \(C=\sqrt{x}+\sqrt{2-x}\)
Đk:\(0\le x\le2\)
\(C^2=x+2-x+2\sqrt{x\left(2-x\right)}\)
\(=2+2\sqrt{x\left(2-x\right)}\) (bđt Cauchy)
\(\le2+x+\left(2-x\right)\)
\(=2+2=4\)
\(\Rightarrow C^2\le4\Leftrightarrow C\le2\)
Dấu = khi \(\sqrt{x}=\sqrt{2-x}\Leftrightarrow x=1\)
Vậy MaxC=2 khi x=1