\(\dfrac{x-1}{x+3}\le x\)

\(\dfrac{x-1}{x+3}-x\le0\)

\(\dfrac{x-1-x\left(x+3\right)}{x+3}\le0\)

\(\dfrac{-\left(x^2+2x+1\right)}{x+3}\le0\)

\(\dfrac{-\left(x+1\right)^2}{x+3}\le0\)

Vì \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow x+3>0\)

\(\Rightarrow x>-3\)

`Answer:`

b) \(\left|x-1\right|-2\left|x\right|=-2\)

Trường hợp 1: \(-\left(x-1\right)-2.\left(-x\right)=-2\) (Với `x<0`)

\(\Leftrightarrow-x+1+2x=-2\)

\(\Leftrightarrow x+1=-2\)

\(\Leftrightarrow x=-3\)

Trường hợp 2: \(-\left(x-1\right)-2x=-2\) (Với `x>=1`)

\(\Leftrightarrow-x+1-2x=-2\)

\(\Leftrightarrow-3x+1=-2\)

\(\Leftrightarrow-3x=-3\)

\(\Leftrightarrow x=1\)

Cuộc chiến chống Pháp: Phan Đình Giót, Tô Vĩnh Diện, Bế Văn Đàn, Trần Can.

em bik từng đấy thôi chị nha

wow đúng rồi

\(\left(4+\dfrac{1}{4}\right)\left(a^2+\dfrac{1}{b+c}\right)\ge\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)^2\)

\(\Rightarrow\sqrt{a^2+\dfrac{1}{b+c}}\ge\dfrac{2}{\sqrt{17}}\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)=\dfrac{1}{\sqrt{17}}\left(4a+\dfrac{1}{\sqrt{b+c}}\right)\)

Tương tự:

\(\sqrt{b^2+\dfrac{1}{a+c}}\ge\dfrac{1}{\sqrt{17}}\left(4b+\dfrac{1}{\sqrt{a+c}}\right)\) ; \(\sqrt{c^2+\dfrac{1}{a+b}}\ge\dfrac{1}{\sqrt{17}}\left(4c+\dfrac{1}{\sqrt{a+b}}\right)\)

Cộng vế:

\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\right)\)

\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)\)

Cũng theo Bunhiacopxki:

\(1.\sqrt{a+b}+1.\sqrt{b+c}+1\sqrt{c+a}\le\sqrt{\left(1+1+1\right)\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{6\left(a+b+c\right)}}\right)\)

\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}\left(a+b+c\right)+\dfrac{a+b+c}{8}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}\right)\) 

\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}.6+3\sqrt[3]{\dfrac{81\left(a+b+c\right)}{32.6\left(a+b+c\right)}}\right)=\dfrac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra khi \(a=b=c=2\)