\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{ab}{c+a}+\frac{ac}{a+b}+\frac{ab}{b+c}+\frac{b^2}{c+a}+\frac{bc}{a+b}\)

\(+\frac{ca}{b+c}+\frac{bc}{c+a}+\frac{c^2}{a+b}=a+b+c\)

\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{c\left(a+b\right)}{a+b}+\frac{a\left(b+c\right)}{b+c}\)

\(+\frac{b\left(c+a\right)}{c+a}=a+b+c\)

\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\left(a+b+c\right)=a+b+c\)

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\left(đpcm\right)\)

hay

mày làm đi

Ta có: \(2x^2+x+1\)

\(=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{1}{2\sqrt{2}}+\frac{1}{8}+\frac{7}{8}\)

\(=\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

\(\frac{\Rightarrow\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{7}{8}}{-2}\le\frac{-7}{16}\)

(Dấu "="\(\Leftrightarrow\sqrt{2}x+\frac{1}{2\sqrt{2}}=0\Leftrightarrow x=\frac{-1}{4}\)

\(D=\frac{2x^2+x+1}{-2}\)

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

\(=\frac{2\left(x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{1}{2}\right)}{-2}\)

\(=\frac{2\left(x+\frac{1}{2}\right)^2+\frac{7}{8}}{-2}\)

Vì \(2\left(x+\frac{1}{2}\right)^2\ge0;\forall x\)

\(\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{7}{8}\ge\frac{7}{8};\forall x\)

\(\Rightarrow\frac{2\left(x+\frac{1}{2}\right)^2+\frac{7}{8}}{-2}\ge\frac{-7}{16};\forall x\)

Dấu'="xảy ra \(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\)

                      \(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(D_{min}=\frac{-7}{16}\)\(\Leftrightarrow x=\frac{-1}{2}\)

lấy búa đập

thuê trộm hoặc thuê tui :)

~.~

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