áp dụng bđt Cosi cho các số a, b, c không âm, ta có:

\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\left(1\right)\\a+c\ge2\sqrt{ac}\left(2\right)\\b+c\ge2\sqrt{bc}\left(3\right)\end{matrix}\right.\)

lấy (1) + (2) + (3) vế theo vế, ta được:

\(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)\)

\(\Leftrightarrow a+b+c\ge\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\) (đpcm)

Bài 4:

a: ĐKXĐ: x>0; x<>4

Ta có: \(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right)\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{2}{\sqrt{x}+2}\)

b: \(A>\frac12\)

=>\(\frac{2}{\sqrt{x}+2}-\frac12>0\)

=>\(\frac{4-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}>0\)

=>\(2-\sqrt{x}>0\)

=>\(\sqrt{x}<2\)

=>0<x<4

c: \(B=\frac73\cdot A=\frac73\cdot\frac{2}{\sqrt{x}+2}=\frac{14}{3\sqrt{x}+6}\)

Để B nguyên thì 14⋮3\(\sqrt{x}+6\)

=>\(3\sqrt{x}+6\in\left\lbrace7;14\right\rbrace\)

=>\(3\sqrt{x}\in\left\lbrace1;8\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace\frac13;\frac83\right\rbrace\)

=>x\(\in\left\lbrace\frac19;\frac{64}{9}\right\rbrace\)

\(\sqrt{\dfrac{9}{7-4\sqrt{3}}}-\sqrt{\dfrac{4}{7+4\sqrt{3}}}\)

\(=\sqrt{\dfrac{9\left(7+4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}-\sqrt{\dfrac{4\left(7-4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}\)

\(=\dfrac{3\sqrt{7+4\sqrt{3}}-2\sqrt{7-4\sqrt{3}}}{\sqrt{49-48}}\)

\(=3\sqrt{4+2.2.\sqrt{3}+3}-2\sqrt{4-2.2.\sqrt{3}+3}\)

\(=3\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=3.\left|2+\sqrt{3}\right|-2.\left|2-\sqrt{3}\right|\)

\(=6+3\sqrt{3}-2+2\sqrt{3}\)

\(=4+5\sqrt{3}\)

Sửa đề: \(\sqrt{7-4\sqrt3}-\sqrt{7+4\sqrt3}\)

\(=\sqrt{4-2\cdot2\cdot\sqrt3+3}-\sqrt{4+2\cdot2\cdot\sqrt3+3}\)

\(=\sqrt{\left(2-\sqrt3\right)^2}-\sqrt{\left(2+\sqrt3\right)^2}\)

\(=2-\sqrt3-2-\sqrt3=-2\sqrt3\)

Điều kiện : \(x\le3\)

Đặt \(3-x=t\left(t\ge0\right)\Rightarrow5-x=2+t\)

Khi này ta có phương trình :

\(\sqrt{t}+\sqrt{t+2}< 2\)

\(\Leftrightarrow t+2\sqrt{t\left(t+2\right)}+t+2< 2\)

\(\Leftrightarrow2t+2\sqrt{t\left(t+2\right)}< 0\)

\(\Leftrightarrow t+\sqrt{t\left(t+2\right)}< 0\) ( vô lí do \(t\ge0\) )

Vậy bất phương trình đã cho vô nghiệm

Lời giải:

Thay vì dấu < thì dấu $\leq$ đúng hơn 

CMR: $(ax+by)^2\leq (a^2+b^2)(x^2+y^2)$

$\Leftrightarrow (a^2+b^2)(x^2+y^2)-(ax+by)^2\geq 0$

$\Leftrightarrow (a^2x^2+a^2y^2+b^2x^2+b^2y^2)-(a^2x^2+b^2y^2+2axby)\geq 0$

$\Leftrightarrow a^2y^2+b^2x^2-2axby\geq 0$

$\Leftrightarrow (ay-bx)^2\geq 0$ (luôn đúng) 

Vậy ta có đpcm.

Dấu "=" xảy ra khi $ay=bx$

bạn thấy đó:\(A^2-B=0\)

nên : Nên theo cái dấu mà bạn viết thì VP=0

bạn thấy vế trái có thể =0 ko?

\(A=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\right)\)

\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{2+\left|\sqrt{3}+1\right|}+\dfrac{2-\sqrt{3}}{2-\left|\sqrt{3}-1\right|}\right)\)

\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\right)\)

\(=\sqrt{2}\left(1-\dfrac{1}{3+\sqrt{3}}+1-\dfrac{1}{3-\sqrt{3}}\right)\)

\(=\sqrt{2}\left(2-\dfrac{3-\sqrt{3}+3+\sqrt{3}}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\right)\)

\(=\sqrt{2}\left(2-\dfrac{6}{9-3}\right)\)

\(=\sqrt{2}\)

\(\dfrac{1}{\sqrt{2}}A=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{2}}{3-\sqrt{3}}=\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6-3\sqrt{2}+2\sqrt{3}-\sqrt{6}}{6}\)

\(=\dfrac{9-3\sqrt{2}-\sqrt{6}+3\sqrt{3}}{6}.\sqrt{2}=\dfrac{9\sqrt{2}-6-2\sqrt{3}+3\sqrt{6}}{6}=\dfrac{3}{2}\sqrt{2}-1-\dfrac{1}{3}\sqrt{3}+\dfrac{1}{2}\sqrt{6}\)