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b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
1) \(\left(a-b\right)^2\ge0\)
\(a^2-2ab+b^2\ge0\)
\(a^2+b^2+2ab\ge4ab\)
\(\left(a+b\right)^2\ge4ab\)
\(\dfrac{\left(a+b\right)^2}{4}\ge ab\)
\(\dfrac{a+b}{2}\ge\sqrt{ab}\)
Dấu ''='' xảy ra khi a=b
2) \(\left(\sqrt{2a}-\sqrt{2b}\right)^2\ge0\)
\(2a-4\sqrt{ab}+2b\ge0\)
\(4a+4b\ge2a+2b+4\sqrt{ab}\)
\(\dfrac{a+b}{2}\ge\dfrac{a+b+2\sqrt{ab}}{4}\)
\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
Dấu ''='' xảy ra khi a=b
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)
\(\Leftrightarrow2n+1=1\left(2n+1\right)\)
\(\Leftrightarrow2n+1=2n+1\)
\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
Câu b) ý 2:
Áp dụng BĐT cô si ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
Ta có \(c\ge\sqrt{ab}\Leftrightarrow c^2\ge ab\Leftrightarrow c^2-ab\ge0\Leftrightarrow c\left(c^2-ab\right)\ge0\Leftrightarrow c^3-abc\ge0\Leftrightarrow\left(c^3-abc\right)\left(a-b\right)\ge0\Leftrightarrow ac^3-a^2bc-bc^3+ab^2c\ge0\Leftrightarrow ab^2c+ac^3\ge a^2bc+bc^3\Leftrightarrow ac\left(b^2+c^2\right)\ge bc\left(a^2+c^2\right)\Leftrightarrow\dfrac{ac}{a^2+c^2}\ge\dfrac{bc}{b^2+c^2}\Leftrightarrow\dfrac{2ac}{a^2+c^2}\ge\dfrac{2bc}{b^2+c^2}\Leftrightarrow1+\dfrac{2ac}{a^2+c^2}\ge1+\dfrac{2bc}{b^2+c^2}\Leftrightarrow\dfrac{a^2+2ac+c^2}{a^2+c^2}\ge\dfrac{b^2+2bc+c^2}{b^2+c^2}\Leftrightarrow\dfrac{\left(a+c\right)^2}{a^2+c^2}\ge\dfrac{\left(b+c\right)^2}{b^2+c^2}\Leftrightarrow\dfrac{a+c}{\sqrt{a^2+c^2}}\ge\dfrac{b+c}{\sqrt{b^2+c^2}}\left(đpcm\right)\)
Cần chứng minh
(a + c)²(b² + c²) ≥ (b + c)²(a² + c²)
<=> 2c(a - b)(c² - ab) ≥ 0
Cái này đúng.
a: \(a+\dfrac{1}{a}\ge2\sqrt{a\cdot\dfrac{1}{a}}=2\)
b: \(\Leftrightarrow\dfrac{a^2+a+1+1}{\sqrt{a^2+a+1}}>=2\)
=>\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\)(1)
\(\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}>=2\sqrt{\sqrt{a^2+a+1}\cdot\dfrac{1}{\sqrt{a^2+a+1}}}=2\)
nên (1) đúng
a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
Lời giải:
Biến đổi tương đương:
\(\sqrt{\frac{a+b}{2}}\geq \frac{\sqrt{a}+\sqrt{b}}{2}\)
\(\Leftrightarrow \frac{a+b}{2}\geq \frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{a+b+2\sqrt{ab}}{4}\)
\(\Leftrightarrow \frac{a+b}{2}-\frac{a+b+2\sqrt{ab}}{4}\geq 0\)
\(\Leftrightarrow \frac{a+b-2\sqrt{ab}}{4}\geq 0\)
\(\Leftrightarrow \frac{(\sqrt{a}-\sqrt{b})^2}{4}\geq 0\) (luôn đúng)
Do đó ta có đpcm
Dấu "=" xảy ra khi $a=b$