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a) ĐKXĐ: \(-2\le x\le2\)
\(M^2=2-x+x+2+2\sqrt{\left(2-x\right)\left(x+2\right)}=4+2\sqrt{\left(2-x\right)\left(x+2\right)}\)
\(\ge4\)\(\Rightarrow M\ge2\) Vậy min M = 2\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)
Mặt khác \(M^2=4+2\sqrt{\left(2-x\right)\left(2+x\right)}\le4+2-x+2+x=8\)
\(\Rightarrow M\le2\sqrt{2}\) Vậy max M = \(2\sqrt{2}\Leftrightarrow x=0\)(thỏa mãn)
Câu b tương tự nhé
ĐKXĐ: ...
\(M\ge\sqrt{2-x+x+2}=2\)
\(M_{min}=2\) khi \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(M\le\sqrt{2\left(2-x+x+2\right)}=2\sqrt{2}\)
\(M_{max}=2\sqrt{2}\) khi \(2-x=x+2\Leftrightarrow x=0\)
\(N\ge\sqrt{x-3+5-x}=\sqrt{2}\)
\(N_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
\(N\le\sqrt{2\left(x-3+5-x\right)}=2\)
\(N_{max}=2\) khi \(x-3=5-x\Leftrightarrow x=4\)
1 ) \(A=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ : \(2\le x\le4\)
\(\Rightarrow A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)
Áp dụng bđt AM - GM ta có :
\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\)
\(\Rightarrow A^2\le2+2=4\Rightarrow-2\le A\le2\)
Mà A > 0 nên ko thể có min = - 2 nên \(2\le x\le4\) ta chọn x = 2
=> A = \(\sqrt{2}\)
Vậy \(\sqrt{2}\le A\le2\)
a: \(A=x+\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=0\)
Dấu '=' xảy ra khi x=0
b: \(B=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}< =-\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/4
c: \(=x-2005-\sqrt{x-2005}+2005\)
\(=\left(\sqrt{x-2005}\right)^2-2\cdot\sqrt{x-2005}\cdot\dfrac{1}{2}+\dfrac{1}{4}+2004.75\)
\(=\left(\sqrt{x-2005}-\dfrac{1}{2}\right)^2+2004.75>=2004.75\)
Dấu '=' xảy ra khi x=2005,25
d: \(D=x-2+2\sqrt{x-2}+2\)
\(=\left(\sqrt{x-2}+1\right)^2+1>=2\)
Dấu '=' xảy ra khi x=2
\(A^2=\left(2\sqrt{x-4}+\sqrt{8-x}\right)^2\le\left(2^2+1^2\right)\left(x-4+8-x\right)=20..\)
\(A\le2\sqrt{5}..\)
1) \(A=\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=3-2\sqrt{2}\)
\(B=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}-1+2-\sqrt{3}=1\)
\(C=\sqrt{63}-\sqrt{28}-\sqrt{7}=3\sqrt{7}-2\sqrt{7}-\sqrt{7}=0\)
\(D=\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}=\frac{4}{2}=2\)
\(M=\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}=\frac{3+\sqrt{5}-3+\sqrt{5}}{9-5}.\frac{\sqrt{5}-1}{\sqrt{5}\left(\sqrt{5}-1\right)}=\frac{2}{4}=\frac{1}{2}\)