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1) \(A=\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=3-2\sqrt{2}\)
\(B=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}-1+2-\sqrt{3}=1\)
\(C=\sqrt{63}-\sqrt{28}-\sqrt{7}=3\sqrt{7}-2\sqrt{7}-\sqrt{7}=0\)
\(D=\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}=\frac{4}{2}=2\)
\(M=\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}=\frac{3+\sqrt{5}-3+\sqrt{5}}{9-5}.\frac{\sqrt{5}-1}{\sqrt{5}\left(\sqrt{5}-1\right)}=\frac{2}{4}=\frac{1}{2}\)
A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
Lời giải:
a) ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$
b) ĐKXĐ: $3+2x>0\Leftrightarrow x>\frac{-3}{2}$
c) ĐKXĐ: $x^2-4\geq 0\Leftrightarrow (x-2)(x+2)\geq 0$
$\Leftrightarrow x\geq 2$ hoặc $x\leq -2$
d)
ĐKXĐ\(\left\{\begin{matrix} x\geq 0\\ \sqrt{x}\neq 2\\ x+1>0\\ x\neq 0\\ \sqrt{x}\neq 3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>0\\ x\neq 4\\ x\neq 9\end{matrix}\right.\)
e)
ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ 7-\sqrt{x}>0\end{matrix}\right.\Leftrightarrow 0\leq x< 49\)
f)
\(\left\{\begin{matrix} 5-x\neq 0\\ \frac{x+3}{5-x}\geq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x+3\geq 0\\ 5-x>0\end{matrix}\right.\\ \left\{\begin{matrix} x+3\leq 0\\ 5-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow -3\leq x< 5\)
Lời giải:
a)
\(\left\{\begin{matrix} x\geq 0\\ 3-\sqrt{x}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\leq 9\end{matrix}\right.\Leftrightarrow 0\leq x\leq 9\)
b)
\(\left\{\begin{matrix} x-1\geq 0\\ 2-\sqrt{x-1}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x-1\leq 4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 5\end{matrix}\right.\)
\(\Leftrightarrow 1\leq x\leq 5\)
c)
\(-7+3x>0\Leftrightarrow x>\frac{7}{3}\)
d)
\(\left\{\begin{matrix} x-1\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x< 5\end{matrix}\right.\Leftrightarrow 1\leq x< 5\)
e) \(x\in\mathbb{R}\)
f) \(\left\{\begin{matrix} 2-x>0\\ x-5\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x< 2\\ x\geq 5\end{matrix}\right.\) (vô lý)
Do đó không tồn tại $x$ để hàm số tồn tại
g)
\(\left[\begin{matrix} \left\{\begin{matrix} 3x-6-2x\geq 0\\ 1-x>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x-6-2x\leq 0\\ 1-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq 6\\ x< 1\end{matrix}\right.(\text{vô lý})\\ \left\{\begin{matrix} x\leq 6\\ x>1 \end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow 1< x\leq 6\)
Lời giải:
ĐK để tồn tại các biểu thức là $x\geq 0$
a) Ta thấy: $\sqrt{x}\geq 0\Rightarrow \sqrt{x}+5\geq 5$
$\Rightarrow A=\frac{2}{\sqrt{x}+5}\leq \frac{2}{5}$
Vậy $A_{\max}=\frac{2}{5}$ khi $x=0$
b) $\sqrt{x}+7\geq 7$
$\Rightarrow \frac{1}{\sqrt{x}+7}\leq \frac{1}{7}$
$\Rightarrow B=\frac{-3}{\sqrt{x}+7}\geq \frac{-3}{7}$
Vậy $B_{\min}=\frac{-3}{7}$ khi $x=0$
c)
$2\sqrt{x}+1\geq 1\Rightarrow C=\frac{5}{2\sqrt{x}+1}\leq 5$
Vậy $C_{\max}=5$ khi $x=0$
d)
$3\sqrt{x}+2\geq 2\Rightarrow \frac{1}{3\sqrt{x}+2}\leq \frac{1}{2}$
$\Rightarrow D=\frac{-7}{3\sqrt{x}+2}\geq \frac{-7}{2}$
Vậy $B_{\min}=\frac{-7}{2}$ khi $x=0$