a. `A>=0`.

A.

mịa c đâu ra vậy

Ta có :

\(a-\sqrt{a}+\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\forall a\ge0\Rightarrow a+\frac{1}{4}\ge\sqrt{a}\)

\(b-\sqrt{b}+\frac{1}{4}=\left(\sqrt{b}-\frac{1}{2}\right)^2\ge0\forall b\ge0\Rightarrow b+\frac{1}{4}\ge\sqrt{b}\)

\(\Rightarrow a+\frac{1}{4}+b+\frac{1}{4}\ge\sqrt{a}+\sqrt{b}\)

\(\Rightarrow a+b+\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\)(đpcm)

Ta có:

\(S=\dfrac{a^2}{a\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{b^2}{b\left(\sqrt{c}+\sqrt{a}\right)}+\dfrac{c^2}{c\left(\sqrt{a}+\sqrt{b}\right)}\ge\dfrac{\left(a+b+c\right)^2}{a\left(\sqrt{b}+\sqrt{c}\right)+b\left(\sqrt{c}+\sqrt{a}\right)+c\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(b+c\right)+\sqrt{b}\left(c+a\right)+\sqrt{c}\left(a+b\right)}\)

Mặt khác:

\(\sqrt{a}\left(b+c\right)=\dfrac{1}{\sqrt{2}}\sqrt{2a.\left(b+c\right)\left(b+c\right)}\le\dfrac{1}{\sqrt{2}}\sqrt{\left(\dfrac{2a+2b+2c}{3}\right)^3}=\dfrac{2\sqrt{3}}{9}\)

\(\Rightarrow S\ge\dfrac{1}{3.\dfrac{2\sqrt{3}}{9}}=\dfrac{\sqrt{3}}{2}\)

a, \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)

\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

\(=-\sqrt{3}\)

b, \(\sqrt{81a}-\sqrt{36a}+\sqrt{144a}\)

\(=9\sqrt{a}-6\sqrt{a}+12\sqrt{a}\)

\(=15\sqrt{a}\)

c, \(\dfrac{4}{\sqrt{5}-2}-\dfrac{4}{\sqrt{5}+2}\)

\(=\dfrac{4\sqrt{5}+8-4\sqrt{5}+8}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{16}{5-4}=16\)

d, \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{ab}\)

 P = A.B = \(\dfrac{x-7}{\sqrt{x}+2}=\dfrac{\left(x-4\right)-3}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)-3}{\sqrt{x}+2}\)

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

\(P\inℤ\) <=> x là số chính phương và \(\dfrac{3}{\sqrt{x}+2}\inℤ\)

mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow x=1\) (thỏa)

Vậy x = 1 thì P \(\inℤ\)

b) Thay x=49 vào A, ta được:

\(A=\dfrac{7-1}{7-5}=\dfrac{6}{2}=3\)

a) Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)

\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)

Lời giải:

Ta có:

$P^2=2+2(a+b)+2\sqrt{(1+2a)(1+2b)}=2+2+2\sqrt{1+2(a+b)+4ab}$

$=4+2\sqrt{3+4ab}$

Vì $a,b\geq 0$ nên $\sqrt{3+4ab}\geq \sqrt{3}$

$\Rightarrow P^2\geq 4+2\sqrt{3}$

$\Rightarrow P\geq \sqrt{3}+1$
Vậy $P_{\min}=\sqrt{3}+1$. Giá trị này được khi $(a,b)=(1,0)$ và hoán vị.