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bài 1
\(K=x^2+x+1=x^2+2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>=\frac{3}{4}\)
dấu = xảy ra khi \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
vậy min của K là 3/4 tại x=-1/2
bài 2
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc=0^2=0\)
\(\Rightarrow2+2ab+2ac+2bc=0\Rightarrow2ab+2ac+2bc=-2\Rightarrow ab+ac+bc=-1\)
\(\left(ab+ac+bc\right)^2=a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\)
\(=a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=a^2b^2+a^2c^2+b^2c^2=\left(-1\right)^2=1\)
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=a^4+b^4+c^4+2=2^2=4\)
\(\Rightarrow a^4+b^4+c^4=2\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2 \Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
<=> \(ab+bc+ac=0\Leftrightarrow\frac{ab+ac+bc}{abc}=0\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\frac{1}{c^3}\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+3.\frac{1}{a^2}.\frac{1}{b}+3.\frac{1}{a}.\frac{1}{b^2}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{-1}{c}\right)=0\Leftrightarrow\)dpcm
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
=\(\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{ac+c+abc}\)
=\(\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)
=\(\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{1}{a+1+ab}\)
=\(\dfrac{a+1}{ab+a+1}+\dfrac{ab}{abc+ab+a}\)
=\(\dfrac{a+abc}{ab+a+abc}+\dfrac{ab}{abc+ab+a}\)
=\(\dfrac{a+abc+ab}{ab+a+abc}=1\) (đpcm)