Nguyễn Tiến Bình cho mình cách làm đi

15

Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có

\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)

\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

                                                         Bài giải

                                            Số nhóm cô giáo đã chia là:

                                                    32:4=8 (nhóm)

                                                        Đáp số:8 nhóm.

                                                                     Đúng không?

Đặt \(\dfrac{x}{y}+\dfrac{y}{x}=a\)\(\Rightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=a^2\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}=a^2-2\)

Ta có \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4=a^2-2+4=a^2+2\)

\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)=3a\)

Ta có \(a^2+2-3a=a^2-2.a.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}=\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)

lạ có \(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2}{xy}-\dfrac{2xy}{xy}+\dfrac{y^2}{xy}=\dfrac{\left(x-y\right)^2}{xy}\ge0\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge2\)\(\Rightarrow a\ge2\Rightarrow a-\dfrac{3}{2}\ge\dfrac{1}{2}\)\(\Rightarrow\left(a-\dfrac{3}{2}\right)^2\ge\dfrac{1}{4}\Rightarrow\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge0\)

\(\Rightarrow a^2+2-3a\ge0\Rightarrow a^2+2\ge3a\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

Gọi tổng trên là A

\(\Leftrightarrow A=1+\left(-2\right)+2^2+...+2^{2014}+\left(-2\right)^{2015}\)

\(\Rightarrow2.A=2.1+\left(-2.2\right)+2.2^2+...+\left(-2.2\right)^{2015}\)

\(\Rightarrow2A=2+\left(-2\right)^2+2^3+...+\left(-2\right)^{2016}\)

\(\Leftrightarrow2A=2+2^2+2^3+...+2^{2016}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left[1+\left(-2\right)+2^3+...+\left(-2\right)^{2015}\right]\)\(\Rightarrow A=2^{2016}-1\)

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

ta có \(9x^2+16=0\)

Với mọi x ta luôn có \(x^2\ge0\Rightarrow9x^2\ge0\)

\(\Rightarrow9x^2+16>0\left(ktm\right)\)

=> ptrình vô nghiệm

vậy tập nghiệm của phương trình là S=\(\varnothing\)