a) \(\sqrt{81}-\sqrt{80}.\sqrt{0,2}=9-4\sqrt{5}.\frac{\sqrt{5}}{5}=9-\frac{4.5}{5}=9-4=5\)

b) \(\sqrt{\left(2-\sqrt{5}\right)^2}-\frac{1}{2}\sqrt{20}=\left|2-\sqrt{5}\right|-\frac{1}{2}.2\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)

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

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

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

\(=3-\sqrt{3}+\sqrt{3}-1\)

\(=2\)

\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)

\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)

\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}=1-\frac{1}{4}=\frac{3}{4}\)

\(P=\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)

\(=1-\frac{1}{3}\left(\frac{c^2}{c^2+3c\sqrt{ab}}+\frac{a^2}{a^2+3a\sqrt{bc}}+\frac{b^2}{b^2+3b\sqrt{ca}}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)

\(\le1-\frac{1}{3}\cdot\left(\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right)\)

\(\le1-\frac{1}{3}\cdot\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}}\)

\(=1-\frac{1}{4}\)

\(=\frac{3}{4}\)

cho A=1+2⁺2^2+.........+2²⁰⁰⁹+2²⁰¹⁰.Tìm số dư khi chia a cho 7

giải phương trình

x=0 hoặc x=-1

a(x-b)(x-c)+b(x-a)(x-c)+c(x-a)(x-b)=0 (*) 
<=> (a+b+c)x^2 -2x(ab+bc+ca) +3abc =0 

D'(Delta ') = (ab+bc+ca)^2 - 3abc(a+b+c) (**) 

Áp dụng BĐT vào (**): (x+y+z)^2/3 >= xy+yz+zx 
<=> D' = (ab+bc+ca)^2 - 3abc(a+b+c) >= 0 

=> Phương trình (*) luôn có nghiệm với mọi a, b, c

Ko chắc nha !

Minh Anh