đay là violympic lớp 5 thì phải

Đề sai rồi!

2. \(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)= \(\sqrt{9-2\cdot3\cdot\sqrt{2}+2}+\sqrt{2-2\sqrt{2}\cdot1+1}\)

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

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

Đặt A=\(\left(a+\frac{1}{a}\right)x^2y^6=\frac{a^2+1}{a}\cdot x^2y^6\)

Ta thấy \(a^2+1>0;x^2y^6\ge0\)  => Để A <0 thì a <0.

a)Ta có: \(A=x^2+5y^2-2xy+4y+3\)= \(\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

                    = \(\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

(Do \(\left(x-y\right)^2\ge0;\left(2y+1\right)^2\ge0\))

Vậy min A=2. Dấu = khi x=y=-1/2

b) Đặt \(t=x^2-2x+1\)

=> \(B=\left(t-1\right)\left(t+1\right)\)=\(t^2-1\)=\(t^2+\left(-1\right)\ge-1\)

Do \(t^2\ge0\)

Vậy min B=-1. Dấu = khi t=0 hay \(x^2-2x+1=0\)

                                          => \(\left(x-1\right)^2=0\)<=> x=1

Ta có: A=\(\frac{1}{10\cdot12}+\frac{1}{12\cdot14}+\frac{1}{14\cdot16}+...+\frac{1}{38\cdot40}\)

=> \(A=\frac{1}{4}\cdot\left(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{19\cdot20}\right)\)

          =\(\frac{1}{4}\cdot\left(\frac{6-5}{5\cdot6}+\frac{7-6}{6\cdot7}+\frac{8-7}{7\cdot8}+...+\frac{20-19}{19\cdot20}\right)\)

         =  \(\frac{1}{4}\cdot\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\right)\)

         = \(\frac{1}{4}\cdot\left(\frac{1}{5}-\frac{1}{20}\right)\)

         = \(\frac{1}{4}\cdot\frac{3}{20}=\frac{3}{80}\)

Vậy A= 3/80

Ta có: \(4x^2-28x+51=\left(2x\right)^2-2\cdot2x\cdot7+49+2\)

                                       \(=\left(2x-7\right)^2+2\)(*)

Vì \(\left(2x-7\right)^2\ge0\) với mọi x

=> (*)\(\ge1\)

 =>(*) luôn luôn dương với mọi x