\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)

\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)

\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)

\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)

\("="\Leftrightarrow x=3\)

\(P=x^4-6x^3+10x^2-6x+9\)

\(P=\left(x^4-6x^3+9x^2\right)+\left(x^2-6x+9\right)\)

\(P=x^2\left(x^2-6x+9\right)+\left(x^2-6x+9\right)=\left(x^2+1\right)\left(x-3\right)^2\ge0\)Dấu "=" xảy ra khi x=3

\(M=\frac{3}{4x^2-4x+5}=\frac{3}{4x^2-4x+1+4}=\frac{3}{\left(2x-1\right)^2+4}\le\frac{3}{4}\)

Dấu "=" xảy ra khi x=\(\frac{1}{2}\)

\(A=\frac{2}{6x-5-9x^2}\Rightarrow-A=\frac{2}{9x^2-6x+5}=\frac{2}{9x^2+6x+1+4}=\frac{2}{\left(3x+1\right)^2+4}\le\frac{1}{2}\Rightarrow A\ge-\frac{1}{2}\)Dấu "=" xảy ra khi \(x=-\frac{1}{3}\)

\(D=\left(x^4-2x^3+x^2\right)+\left(2x^2-2x+1\right)\)

\(D=\left(x^2-x\right)^2+2\left(x^2-x\right)+1=\left(x^2-x+1\right)^2\)

\(D=\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]^2\)

\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow D\ge\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)

đẳng thúc khi x=1/2

{logic 10x-->10x^2}

\(E=x^4-6x^3+10x^2-6x+9\)

\(E=\left(x^4-3x+9x^2\right)+\left(x^2-6x+9\right)\)

\(E=\left(x^2-3x\right)^2+\left(x-3\right)^2=\left[x^2\left(x-3\right)^2\right]+\left(x-3\right)^2\)

\(E=\left(x-3\right)^2\left(x^2+1\right)\)

\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(x^2+1\right)\ge1\end{matrix}\right.\) \(\Rightarrow E\ge0\) đẳng thức khi x=3

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI