M = x⁴ - 6x³ + 10x² - 6x + 9

M = (x² - 6x + 9) + x⁴ - 6x³ + 9x²

M = (x - 3)² + x²(x² - 6x + 9)

M = (x - 3)².(1 + x²)

Ta có:\(\left(x-3\right)^2\ge0;\left(1+x^2\right)\ge1\)

\(\Rightarrow M\ge1\)

Dấu 'x' xảy ra khi:

\(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Mmin = 1 khi x = 3

Chúc bạn học tốt!!!

Mình giải lại từ dòng số 6 nhé!!!

=> M = 0 

Dấu '=' xảy ra khi:

(x - 3)² = 0 => x - 3 = 0

=> x = 3

Vậy Mmin = 0 khi x = 3

\(\sqrt{x^2+6x+9}+\sqrt{x^2-4x+4}\)

= |x + 3| + |x - 2|

= |x + 3| + |2 - x| \(\ge\)|x + 3 + 2 - x| = 5

Vậy GTNN của M = 5

chưa vội kết luận nha, nãy ghi lộn:

Dấu "=" xảy ra khi <=> (x + 3)(2 - x) >= 0 (tự giải ra)

Vậy GTNN của M bằng 5 khi ....

Bài như vậy đúng ko ạ 

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

\(M=\sqrt{\left(x+3\right)^2}+\sqrt{\left(x-2\right)^2}\)

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

Có GTNN của \(M=\left|x+3\right|+\left|x-2\right|\ge x+3+2-x\)

                                                                          \(\ge5\)

=> GTNN của \(M=\sqrt{x^2+6x+9}+\sqrt{x^2-4x+4}\)là 5

            Tại \(x+3=0\)

                  \(2-x=0\)

               \(\Rightarrow x\in\left\{-3;2\right\}\)

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

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

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

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

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

MinC = 3/4 (khi x = -3/2)

\(C=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4+3x^2-3x^3-9x+3x^2+9}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ C=\dfrac{\left(x^2+3\right)^2-9x^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}\\ C=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)

\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)

Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)

\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)

\(\left(1\right)\left(2\right)\Leftrightarrow B_{min}=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(C=2x^2+4x+2+3x^2+12x+12-4x^2-24x-36\\ C=x^2-8x-22=\left(x^2-8x+16\right)-38=\left(x-4\right)^2-38\ge-38\\ C_{min}=-38\Leftrightarrow x=4\)

Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)

\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)

Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)

\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)