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C=\(\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{\left(x^4+3x^2\right)-\left(3x^3+9x\right)+\left(3x^2+9\right)}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3x\right)\left(x^2-3x+3\right)}=\dfrac{\left(x^2+3+3x\right)\left(x^2+3-3x\right)}{x^2+3-3x}=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}+3=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) Dấu "=" xảy ra \(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy Min C bằng \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{-3}{2}\)
\(A=3x^2-12x+10\\ A=3x^2-12x+12-2\\ A=\left(3x^2-12x+12\right)-2\\ A=3\left(x^2-4x+4\right)-2\\ A=3\left(x^2-2\cdot x\cdot2+2^2\right)-2\\ A=3\left(x-2\right)^2-2\\ Do\left(x-2\right)^2\ge0\forall x\\ \Rightarrow3\left(x-2\right)^2\ge0\forall x\\ \Rightarrow A=3\left(x-2\right)^2-2\ge-2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{ Vậy }A_{\left(Min\right)}=-2\text{ khi }x=2\)
A=3x2 - 12x + 10
A= (3x2- 2.3x.2+22)-22+10
A= (3x-2)2+6 \(\ge\) +6
Vậy min A = 6 . Dấu = xảy ra khi 3x -2 = 0
3x= 2
x= \(\dfrac{2}{3}\)
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
Vì x,y,z >0
Áp dụng BĐT Cosy:
\(x+y\ge2\sqrt{xy}\) (1)
\(x+z\ge2\sqrt{xz}\) (2)
\(y+z\ge2\sqrt{yz}\) (3)
Nhân 3 vế ta được:
\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{xz}\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8xyz\left(đpcm\right)\)
Chúc bạn học tốt 
áp dụng BĐT cauchy, ta có:
\(\left[{}\begin{matrix}x+y\ge2\sqrt{xy}\\y+z\ge2\sqrt{yz}\\x+z\ge2\sqrt{xz}\end{matrix}\right.\)
nhân vế theo vế các BĐT trên, ta được:
\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{xz}=8\sqrt{\left(xyz\right)^2}=8xyz\)
Ta có: \(-x^2+2x-3=-x^2+2x-1-2=-\left(x-1\right)^2-2\le-2\) (1)
Và \(A=\dfrac{-5}{x^2-2x+3}=\dfrac{5}{-x^2+2x+3}\) (2)
Từ (1);(2)\(\Rightarrow A\ge-\dfrac{5}{2}\) Vậy min A=-5/2 khi x=1