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a) Bình phương 2 vế được: \(\frac{4ab}{a+b+2\sqrt{ab}}\le\sqrt{ab}\)
<=> \(4ab\le\sqrt{ab}\left(a+b\right)+2ab\)
<=>\(\sqrt{ab}\left(a+b\right)\ge2ab\)
<=>\(a+b\ge2\sqrt{ab}\)
<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
Vậy \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\forall a,b>0\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b/ Ko biết yêu cầu
4/ \(E=\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{x^3}+\frac{1}{x^3}\ge5\sqrt[5]{\frac{x^6}{27x^6}}=\frac{5}{\sqrt[5]{27}}\)
Dấu "=" xảy ra khi \(\frac{x^2}{3}=\frac{1}{x^3}\Leftrightarrow x=\sqrt[5]{3}\)
\(F=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge3\sqrt[3]{\frac{x^2}{4x^2}}=\frac{3}{\sqrt[3]{4}}\)
Dấu "=" xảy ra khi \(\frac{x}{2}=\frac{1}{x^2}\Rightarrow x=\sqrt[3]{2}\)
6/ \(Q=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{8\left(x+1\right)}{2\left(x+1\right)}}=4\)
Dấu "=" xảy ra khi \(\frac{x+1}{2}=\frac{8}{x+1}\Leftrightarrow x=3\)
7/
\(R=\frac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\frac{25\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}=10\)
Dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow x=4\)
8/
\(S=x^2+\frac{2000}{x}=x^2+\frac{1000}{x}+\frac{1000}{x}\ge3\sqrt[3]{\frac{1000^2x^2}{x^2}}=300\)
Dấu "=" xảy ra khi \(x^2=\frac{1000}{x}\Leftrightarrow x=10\)
a.
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\left(x-\sqrt{x}-2\sqrt{x}+2\right)\\ =\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\right]\\ =\frac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\right]\\ =\frac{\sqrt{x}-1}{\sqrt{x}}\)
b.
\(A=\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}}{2\sqrt{x}}< 0\\ \Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\\ \Leftrightarrow x< 4\)
Vậy với 0<x<4 thì A < \(\frac{1}{2}\)
c. Ta có \(A=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)
Để A đạt giá trị nguyên thì \(1⋮\sqrt{x}\Leftrightarrow\sqrt{x}\inƯ\left(1\right)\)
Mà \(\sqrt{x}>0\forall x>0\Rightarrow x=1\)
Vậy với x=1 thì A đạt giá trị nguyên
1,\(x^2-8x+4m^2=0\)
\(\Delta=\left(-8\right)^2-4.4m^2=64-16m^2\)
Để pt có hai nghiệm p/biệt <=> \(\Delta>0\) <=> \(64-16m^2>0\) <=> \(m^2< \frac{64}{16}=4\)
<=>\(-2< m< 2\)
=>\(\sqrt{\Delta}=\sqrt{64-16m^2}=4\sqrt{4-m^2}\)
\(x_1=\frac{8+4\sqrt{4-m^2}}{2}=4+2\sqrt{4-m^2}\)
\(x_2=\frac{8-4\sqrt{4-m^2}}{2}=4-2\sqrt{4-m^2}\)
\(A=\frac{1}{11.m.n}.m.n.\sqrt{\frac{121.m^2}{n^6}}=\frac{1}{11}.\frac{11.m}{n^3}=\frac{m}{n^3}\)
\(B=2\left(m+n\right).\sqrt{\frac{1}{m^2+2mn+n^2}}=2\left(m+n\right).\sqrt{\frac{1}{\left(m+n\right)^2}}=2\)