trang cho oi dau bdt sai roi ^.^ 

tao giai cho may ne 

\(\frac{3+a}{3-a}+\frac{3+b}{3-b}+\frac{3+c}{3-c}\)=\(\frac{2a+b+c}{b+c}+\frac{2b+a+c}{a+c}+\frac{2c+a+b}{a+b}\)

=\(2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+3\ge2\cdot\frac{3}{2}+3=6\)

đến đây tự làm nhé

BĐT Sai kìa.

(a,b,c)=(1;0.5;1,5) 
 

\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

   \(=\sqrt{\frac{2\left(4+\sqrt{7}\right)}{2}}-\sqrt{\frac{2\left(4-\sqrt{7}\right)}{2}}\)

   \(=\sqrt{\frac{8+2\sqrt{7}}{2}}-\sqrt{\frac{8-2\sqrt{7}}{2}}\)

   \(=\sqrt{\frac{7+2\sqrt{7}+1}{2}}-\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)

   \(=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}\)

   \(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)

   \(=\frac{|\sqrt{7}+1|}{\sqrt{2}}-\frac{|\sqrt{7}-1|}{\sqrt{2}}\)

   \(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}\)

   \(=\frac{2}{\sqrt{2}}\)

cho mình hỏi tại sao chia 2 vậy?

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-1=9\)

rtrrfgfffjhfhfd

\(x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

\(\Leftrightarrow x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3\sqrt[3]{\sqrt[3]{\frac{1}{4-\sqrt{5}}}.\sqrt[3]{4-\sqrt{5}}}.x\)

\(=4+\sqrt{15}+4-\sqrt{15}+3x=8+3x\)

=>y=3x+8-3x+1987

=1995

a. ĐK \(\hept{\begin{cases}a\ge0\\a\ne4\\a\ne9\end{cases}}\)

P=\(\frac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\frac{2\sqrt{a}-9-a+9+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)

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

b. P = \(\frac{\sqrt{a}+1}{\sqrt{a}-3}=1+\frac{4}{\sqrt{a}-3}\)

P nguyên \(\sqrt{a}-3\inƯ\left(4\right)\Rightarrow\sqrt{a}-3\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{1;2;4;5;7\right\}\Rightarrow a\in\left\{1;4;16;25;49\right\}\)

c. \(P< 1\Rightarrow P-1< 0\Rightarrow\frac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\Rightarrow\frac{4}{\sqrt{a}-3}< 0\)

\(\Rightarrow0\le a< 9\)và \(a\ne4\)