Giải:

Ta có BĐT phụ: \(\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\le abc\)

Áp dụng BĐT Cauchy - Schwarz ta có:

\(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\)

\(\ge3\sqrt[3]{\dfrac{abc}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)

\(\ge3\sqrt[3]{\dfrac{abc}{abc}}\ge3\) (Đpcm)

Giải:

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(C=\left|x-\dfrac{2}{5}\right|+\left|x-\dfrac{3}{5}\right|=\left|x-\dfrac{2}{5}\right|+\left|\dfrac{3}{5}-x\right|\)

\(\ge\left|x-\dfrac{2}{5}+\dfrac{3}{5}-x\right|=\left|\dfrac{1}{5}\right|=\dfrac{1}{5}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{2}{5}\ge0\\\frac{3}{5}-x\ge0\end{cases}}\) \(\Leftrightarrow\dfrac{2}{5}\le x\le\dfrac{3}{5}\)

Vậy \(C_{min}=\dfrac{1}{5}\) \(\Leftrightarrow\dfrac{2}{5}\le x\le\dfrac{3}{5}\)

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Giải:

Ta có: \(7a+3b⋮23\Rightarrow6\left(7a+3b\right)⋮23\)

\(\Rightarrow6\left(7a+3b\right)+\left(4a+5b\right)⋮23\)

\(\Rightarrow46a+23b⋮23\Rightarrow23\left(2a+b\right)⋮23\) (Đúng)

Vậy \(4a+5b⋮23\) (Đpcm)

Giải:

Phương trình:

\(\sqrt{2010-x}+\sqrt{x-2008}\) \(=x^2-4018x+4036083\) \((*)\)

ĐKXĐ: \(\begin{cases}2010-x \geq 0\\x-2008 \geq 0\end{cases} \) \(\Leftrightarrow2008\le x\le2010\)

Áp dụng BĐT \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\) \(\forall a,b\) ta có:

\(\left(\sqrt{2010-x}+\sqrt{x-2008}\right)^2\) \(\le2\left(2010-x+x-2008\right)\) \(=4\)

\(\Rightarrow\sqrt{2010-x}+\sqrt{x-2008}\le2\left(1\right)\)

Mặt khác:

\(x^2-4018x+4036083=\left(x-2009\right)^2+2\ge2\left(2\right)\)

Từ \(\begin{cases}(1)\\(2)\end{cases} \) suy ra \((*)\) \(\Leftrightarrow VP=VT=2\)

\(\Leftrightarrow\left(x-2009\right)^2=0\Leftrightarrow x-2009=0\Leftrightarrow x=2009\)

Vậy phương trình có 1 nghiệm duy nhất là \(x=2009\)

Giải:

Ta có: \(P=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{abc}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{a+b+c}{abc}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\in Q\) (Đpcm)