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ĐKXĐ \(2\le x\le4\).Đặt A=\(\sqrt[4]{\left(x-2\right)\left(4-x\right)}+\sqrt[4]{x-2}+\sqrt[4]{4-x}+6x\sqrt{3x}\)
Do x\(\ge2>0\)nên ADBĐT CAUCHY ta được:
\(\sqrt[4]{1\cdot1\cdot\left(x-2\right)\left(4-x\right)}\le\frac{1+1+x-2+4-x}{4}=1\)
\(\sqrt[4]{x-2}\le\frac{1+1+1+x-2}{4}=\frac{1}{4}\)
\(\sqrt[4]{4-x}\le\frac{1+1+1+4-x}{4}=\frac{7}{4}\)
\(6x\sqrt{3x}=2\sqrt{27x^3}\le x^3+27\)
_Do đó A\(\le1+\frac{1}{4}+\frac{7}{4}+x^3+27=x^3+30\)
Dấu = xảy ra \(\Leftrightarrow x=3\)(thỏa mãn ĐKXĐ)
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
1. \(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|=1\)
+ Ta có : \(\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)
Dấu "=" \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x}\le3\Leftrightarrow4\le x\le9\)
2. + \(ĐK:4-2x-x^2\ge0\)
+ VT = \(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}\)
\(=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\) \(\ge\sqrt{4}+\sqrt{9}=5\) (1)
Dấu "=" \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
+ VP \(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\forall x\) (2)
Dấu "=" \(\Leftrightarrow x=-1\)
+ Từ (1) và (2) suy ra : pt \(\Leftrightarrow VT=VP=5\Leftrightarrow x=-1\) (TM)
3. + TH1: \(x< 0\) ta có :
\(VT< \sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
+ TH2 : x = 0 ta có :
\(VT=\sqrt[3]{1}+\sqrt[3]{0}=1\) ( TM )
+ TH3 : x > 0 ta có :
\(VT>\sqrt[3]{2.0+1}+\sqrt[3]{0}=1\) ( KTM )
Vậy x = 0 là nghiệm duy nhất của pt
4. \(\Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+3\right)-24=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x-8\right)-24=0\)
\(\Leftrightarrow t\left(t-5\right)-24=0\) ( với \(t=x^2+2x-3\) )
\(\Leftrightarrow t^2-5t-24=0\Leftrightarrow\left(t+3\right)\left(t-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=-3\\x^2+2x-3=8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+2\right)=0\\\left(x+1\right)^2=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\sqrt{3}-1\\x=-2\sqrt{3}-1\end{matrix}\right.\) ( TM )